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W. A. J. CHAPMAN
Senior Workshop
Calculations
Third Edition
SI Units
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Senior workshop calculations
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Senior workshop
calculations
Dr. W. A. J. Chapman
MSc(Eng), FIMechE, HonFIProdE
<D
EDWARD ARNOLD
©W.A.J. Chapman 1972
First published 1941
by Edward Arnold (Publishers) Limited
25 Hill Street
London WIX 8LL
Reprinted 1944, 1947, 1949, 1952
Second edition 1954
Reprinted 1957, 1959, 1961, 1962, 1965
Third edition 1972
Reprinted 1973, 1978
ISBN 7131 3260 4
All Rights Reserved. No part of this publication may be reproduced,
stored in a retrieval system, or transmitted, in any form or by any
means, electronic, mechanical, photocopying, recording or other
wise, without the prior permission of Edward Arnold (Publishers)
Ltd
Photoset and printed in Malta by Interprint (Malta) Ltd
This book is a revision of the workshop practice material in Senior
Workshop Calculations . This revision was made necessary by the advent
of the SI System and of developments in some aspects of teaching and
industrial practice.
Senior Workshop Calculations has enjoyed a vogue of thirty years and
is used in many parts of the world. From the various expressions I
have received I am sure that it has rendered a useful contribution to
the work of countless students as well as to that of mature workers
engaged in engineering.
During the revision it was decided to separate out the text dealing
with practice calculations as a first priority. The text includes most of
the material required by students and practitioners in workshop and
production engineering practice and should provide a useful textbook
for National Certificates and Diplomas, and for City & Guilds courses
in Mechanical and Production Engineering subjects.
In the work of revision I have received considerable advice and help
from Mr M. G. Page, BSc(Eng), FIMechE, FIProdE, and I should like
here to acknowledge my warm appreciation of his generous and kind
assistance.
I hope that in its present form the book will continue to serve the
interests of students and other workers in those aspects of engineering
it has always been my desire to foster.
Hatfield, 1972 W. A. J. C.
Contents
1 Introduction: The SI system of units and conventions
Basic SI units  Supplementary and derived units 
Multiples and subdivisions of the unit  Representation of
unit symbols and quantities  Dimensioning  Mass,
weight and force  The kilogramme weight  Equations of
motion  Acceleration 1
2 Measurements and gauging
Limits and tolerances  Limit systems  Calculations of
limits  Slip gauges  The spirit level  The sine bar 
Gauging large radii and holes  Location of points on
angular surfaces  Measurement of tapers with balls and
rollers  3wire measurement of screw threads 
Miscellaneous measurement problems 13
3 Calculations for cutting, turning and boring
Speeds and feeds  Arithmetic and geometric progression 
Cutting tool life  Tool angles  Effect of tool height 
Taper turning  True shape of form tools, flat and circular 
Approximate change wheels by continued fractions 
Cutting power for turning and drilling 56
4 Calculations for gears and gearcutting
Involute gear teeth  The tooth Vernier  Constant chord
measurement  Module pitch  Plug Method of checking
gear teeth  Base pitch  Stub teeth  Backlash  Helical
(spiral) gear calculations  Worm gearing  Bevel gearing 84
5 Milling and the milling machine
Milling cutters  Cutter teeth  Rake  Clearance  Tooth
angle  Tooth grinding  Helical teeth  Speeds and feeds
for milling cutters  Milling power calculations  The
dividing head  Simple, compound, differential, angular
indexing  Spiral milling  Cam milling  Solid angles 113
6 Mechanical principles I
Vectorial representation  Addition and subtraction of
vectors  Application of vectors forces on a cutting tool 
Balancing of faceplates  Vector velocity diagrams for
mechanisms  Moment of a force  Parallel forces 148
7 Mechanical principles II
Friction  Machines and efficiency  The inclined plane
and screw  Bearings  Bearing pressure  Bearing friction
 Stress and strain 176
8 Mechanical principles III
Velocity  Acceleration  Equations of motion  Force 
Energy; kinetic and potential  Circular motion  Accelerating
torque  Energy of flywheels  Temperature  Heat  Amount
of heat  Heat energy 198
Appendices I and II: ISO standard holes and shafts 2256
Appendices III and IV: BSI standard holes and shafts 2267
Appendix V: Conversion table 2289
Appendix VI: The trigonometrical addition formulae 230
Appendix VII : Continued fractions 23 1
Answers 235
Index 242
1 Introduction
The SI system of units and conventions
The initials SI are an abbreviation for Systeme International d'Unites
(International System of Units), the modern form of the metric system,
finally agreed upon at an international conference in 1960. It is now
being adopted widely throughout the world and is likely to become the
primary world system for units and measurement. As we shall discuss
below, the system rationalises the main metric units of measurement and
standardises their names and symbolic representation. It also rationalises
certain mechanical principles and conventions.
The British system of weights and measures is many centuries old, and
the derivation of its units with their multiples and subdivisions is often
obscure. The system has been refurbished from time to time but the yard
and the pound with their multiples (e.g. mile) and subdivisions (e.g.
ounce) have persisted; so have such ridiculous measures as 5 yards = 1
rod, pole or perch, 4 roods = 1 acre or 141b = 1 stone remained with us to
try the mental agility of generations of students, not to mention the more
mature, and less mentally agile population.
The metric system was founded during the French Revolution and has
been adopted for use by most countries with the notable exceptions of the
British Commonwealth and the U.S.A., but even in these countries it is
used for precise scientific measurements.
The basic units of the SI metric system are the metre and the kilo
gramme and it is exclusively decimal, so that all multiples and sub
divisions of the standard are found by applying factors of 10 (1 kilometre =
1000 metres; 1 kilogramme = 1000 grammes; 1 hectare (area) = 100 x 100
square metres, and so on) . In the English system, of course, there is no such
orderly pattern and indeed, the foreigner might well question our sanity
when he hears us refer to 1 12 pounds as a hundredweight.
However, we have, at last, been caught up with the progress of the times
and as a nation we have decided to change over to the metric system. The
entire text of this book conforms with the SI system and the object of this
introductory chapter is to provide help and reference for the reader as he
finds his way into what may seem, at first, to be a complexity. The best
2 INTRODUCTION
advice that can be given for achieving rapid progress in coping with the
change is to become familiar with the new measures and to learn, as soon
as possible to think in terms of them, and not to persist in making mental
conversions back to the old units. This process can be speeded up by
acquiring, as soon as possible, a mental appreciation of a range of lengths,
weights, capacities, etc. Some of these could be (say) the mental judge
ment of 25 millimetres, 1 metre, the weight of 1 kilogramme, the amount
of fluid comprising 1 litre (1000 cm 3 ) and the pressure corresponding to
1 bar (approx 1 atmosphere)* . In this way it will soon be possible to think
of these measures in their own right and not grope around converting
them to their English equivalents (1kg = 221b; 1 litre = about If pints
and so on). A similar process is concerned in the learning of a foreign
language where fluency will never be achieved until a student thinks in
terms of the language concerned and abandons all attempts to interpret
mentally from, and into, English. It is well known that another language
is quickest learned by living amongst those who speak nothing else. If the
reader can approach this new mathematical and scientific language in
this frame of mind he will find that the former system will rapidly re
cede, so increasing the ease with which he can cope with the problems
involved.
Basic SI units
The SI system is based on six primary units as follows:
Table 1.
Basic SI units
Quantity
Unit
Symbol
Length
Metre
m
Mass
Kilogramme
kg
Time
Second
s
Electric current
Ampere
A
Temperature
Kelvin
K
Luminous Intensity
Candela
cd
In addition to these there are a number of supplementary and derived
units. We give below a selection of those which are most likely to be
required by students using this book. A full list of the SI Units and con
ventions is given in BSI publication PD 5686.
*The reader has probably heard of the bar in reference to millibars of atmospheric
pressure in Met Office forecasts.
SUPPLEMENTARY AND DERIVED SI UNITS
Supplementary and derived units
Quantity
Table 2. Selected supplementary and derived SI units
Unit
Symbol
Area
Square metre
m 2
Volume
Cubic metre
m 3
Density
Kilogramme per cubic metre
kg/m 3
Velocity
Metre per second
m/s
Acceleration
Metre per second squared
m/s 2
Force
Newton
N(kgm/s 2 )
Moment of force
Newton metre
Nm
Pressure, stress
Newton per square metre
N/m 2
Work, energy
Joule
J(Nm)
Heat quantity
Power
Watt
W (J/s)
Plane angle
Radian
rad
Temperature
(Everyday use)
Degree Celsius*
°C
Specific heat capacity
Joule per kilogramme
degree Celsius
J/kg°C
Electric tension )
Potential difference >
Volt
V
Electromotive force )
*It is probable that the word "centigrade" will remain, but SI recommends the use of
Celsius to prevent confusion with another unit.
Multiples and subdivisions of the unit
In the same way, that in the British system where the yard is divided into
feet and inches and multiplied into the furlong and mile, the pound multi
plied into the cwt and ton and divided into the ounce, so it is necessary to
make similar provisions in the SI system. One of the advantages of the
system for this purpose is the simplicity of its multiples and submultiples
because of its decimal (10) character. We will give, again, a selection of the
chief factors likely to be required by the reader leaving him to study the
BSI literature if he wishes to pursue the remainder.
It will be noticed that multiples and divisions involving 10 4 , 10 5 and
certain higher powers are not included in the system. This has been done
in order to rationalise the procedure by using, after 10 2 and 10~ 2 , only
powers which are multiples of 3, and the full list of such factors extends
from 10 12 at the higher end to 10~ 18 at the lower.
INTRODUCTION
Table 3. SI multiples and sub multiples
Factor by which
unit is multiplied
Prefix
Symbol
Example
One million = 10 6
mega
M
megawatt (MW)
One thousand = 10 3
kilo
k
kilometre (km)
One hundred = 10 2
hecto
h
hectare (ha)
Ten = 10
deca
da
decagram (dag)
One tenth = 10'
deci
d
decimetre (dm)
One hundredth = 10 ~ 2
centi
c
centimetre (cm)
One thousandth = 10 ~ 3
milli
m
millilitre (ml)
One millionth = 10  6
micro
/"
microvolt (juV)
Having now considered the quantitative manipulation of the basic
units we may now go on to consider the facilities available for alternative
smaller divisions, or larger multiples of the main units, and of certain
additional variations allowable in the system. By this means it is possible
to employ a unit of suitable proportions for any particular set of circum
stances. (Table 4).
Representation of unit symbols and quantities
It often happens that when we move into a new house, or office, or
take on something which changes our way of life we take the opportunity
of overhauling our methods. So it is with the changeover to this new
system where the adoption of SI is accompanied by various conventions
regarding the presentation of information.
These are summarised as follows:
(a) Writing the unit symbols
(i) The full stop, usually placed after an abbreviated word, is never
used after an abbreviated symbol except at the end of a sentence.
Thus, 50kg, 10m, 8 kg, 30s, 5N/mm 2 , all without a full stop, but:
"the vehicle had a mass of 1100 kg." end of sentence, full stop.
(ii) The plural V is not used. (50cm not 50 cms)
(iii) The proposition "per" is replaced by the oblique / (rev/min
not rev per min)
(iv) Symbols for areas and volumes are qualified by the index, (cm 2
not sq cm and cm 3 not cu cm)
(Note that when converting quantities denoted by indices: lm 3 =
(100cm) 3 = lOCFcm 3 = 10 6 cm 3 ) (continued on p. 6).
REPRESENTATION OF UNIT SYMBOLS AND QUANTITIES
Table 4. Recommended multiples and subdivisions of the basic SI units
Quantity
Plane angle
Length
Area
Volume
Time
Velocity
Mass
Other units acceptable
Multiple or
in the system and
submultiple
likely to persist
m rad
degree minute' second"
km
cm*
mm
/urn
micron (i4mm)
km 2
hectare (ha)(10 4 m 2 )
cm 2
are (a)(10 2 m 2 )
mm 2
dm 3
litre (/)(100 cm 3 )
cm 3
millilitre (ml)
mm 3
Ms
year, month, week.
ks
day (d) hour (h)
ms
minute (min)
/US
Mg
km/h
tonne (OR metric ton) (t)
metric carat (2 x 10  4 kg)
mg
Density
kg/ dm 3
or kg/1
g/cm 3
g/1
Force
MN : kN : mN : pti
kgf (weight of 1 kg mass)
(not included in SI system
but likely to be used)
1 kgf = 9806N
Moment of force
MNm:kNm:,uNm
Pressure
N/mm 2 : N/cm 2 : kN/mm 2
bar = 10N/cm 2
mN/m 2 :/xN/m 2
hectobar (hbar) = 10 3 N/cm 2
(Will probably be used for
Stress
kN/mm 2 : kN/cm 2 : N/mm 2
pneumatic pressures. High
N/cm 2 :kN/m 2
pressures and stresses will
be expressed in units shown
opposite)
Work and energy
MJ:kJ:mJ
Kilowatt hour
(kWh = 36 MJ)
Power
MW:kW:mW:/iW
Specific heat capacity
kJ/kg°C
* The cm is not recommended for general use and it is hoped that it will eventually
disappear. It is, however, a very convenient unit for certain purposes as will be seen
from the uses of it later in our text. It will be observed, however, that the cm 2 and
cm 3 are still permissible.
6 INTRODUCTION
(b) Numerical values
(i) When a quantity is less than unity ( 1 ) always place zero (0) before
the decimal point. (0625:0.0031, etc.)
(ii) As far as possible always express a quantity in terms of a power
of 10, so using the index of 10 instead of a row of 0's. (362 x 10 3
instead of 36 200; 15 x 10~ 3 instead of 00015) and preferably
use powers of 10 which are multiples of 3.
(hi) Separate a row of digits into groups of three by a space, instead
of using a comma. (71 562 instead of 71,562, or 0006 13 instead
of 0006, 1 3) . But a group of four digits may be left without separa
tion (e.g. 6713 or 00036 without separation).
Dimensioning (drawings)
It is customary, when dimensioning drawings, to dimension all sizes
in millimetres, and not to write the unit (mm) after the dimension. An
instruction may be given to the effect that all sizes are in millimetres
but this is not always done. All the diagrams in this book are dimensioned
according to this rule so that the reader will now recognise that <^^ >
means 65 mm.
Mass, weight and force
The reader will observe that kg is the SI unit for MASS and that a unit for
weight is not mentioned in the scheme. There is, however, the Newton (N)
as the unit of FORCE. The mass of a body is the amount of matter (or
material) of which it is composed and the kilogramme unit is equal to
the mass of the international prototype kilogramme which is in the
custody of the Bureau International des Poids et Mesures at Sevres near
Paris. The only means of measuring and comparing masses is by weighing
them so that the weight of an object is proportional to its mass and
indeed, the weight of a body is the downward force its mass exerts under
the influence of the earth's gravitational pull. We have seen this concept
very effectively illustrated since astronauts have been penetrating beyond
the influence of the gravitational pull of the earth and most readers will
have seen the fascinating television pictures of the interior of the capsule,
with objects floating about, still having their mass, but without weight.
The definition of a newton (N) of force is that force which acting on one
kilogramme of mass will propel it along with an acceleration of one metre
per second per second (i.e. it gains 1 m per second every second). Put
into symbols this becomes:
IN = lkgm/s 2
THE KILOGRAMME WEIGHT (kgf) 7
The kilogramme weight (kgf)
The gravitational pull where we are (England) is such that it imparts an
acceleration of about 9807 metres per second per second (9807 m/s 2 )
on a freely falling body and this pull, acting on a mass of 1 kg causes it to
exert a downwards force (its weight) of 9807 Newton of force (since
1 Newton = 1 kgm/s 2 ). Thus the weight of a mass of 1 kg expressed in
newtons of force is:
lkg weight (kgf) = 9807 newtons (usually approximated to 981 N)
The weight of a body having a mass of 150 kg = 150 x 981 = 14715 N)
The mass of a body is a quantity which never changes (unless a piece
is cut off or added to it), but the force of gravity varies slightly on dif
ferent parts of the earth. This means that the weight of 1 kg of mass in
England will not be quite the same as its weight at the equator. The
variation is very small (about 05%) over the surface of the earth so that for
all but the very accurate scientific work there is a justification for using
the kgf unit of weight in everyday life. The reader will find kg loosely
referred to as the "weight" of an article, when what is really meant is
the gravitational pull on a kg of mass. However, for some problems, such
as those involving weights and costs of materials, it will be convenient
to work in kgf rather than newtons, since suppliers of materials will
always quote kg or tonne, and not newtons as the weight in their price
lists and specifications. The following examples will illustrate the use
of force units:
Example 1. A round steel bar, 100 mm diameter, 1 metre long is placed
standing on its end. If the contact between the end of the bar and its
support is uniformly spread over the whole end face of the bar calculate
the intensity of pressure over the area of contact. Take the density of
steel as 783 g/cm 3 .
Mass of bar = (Volume)(density)
= j X 10 2 x 100 x 783 grams (Working in cm)
07854 x 10 4 x 783.
= __ kg = 615 kg
Downward force (weight) exerted by bar
= (mass) (gravity) === 615 X 9807 = 603 N
Area of bar end = 07854 x 10 2 cm 2
= 7854 cm 2
INTRODUCTION
T . c force 603
Intensity of pressure =
area 7854
= 768 N/cm 2
Example 2. A vehicle having a mass (m) of 2000 kg has a fractional
resistance to motion of m/20. If the drive is equivalent to an average
force of 320 N find the speed attained after 1 min from rest.
2000
Resistance to motion = ^— N = 100 N
Effective driving force = 320N  100N = 220N
and since force (F) = mass (m) x accel (a)
220N = 2000 a
220 All ,2
" = 2000 =° llm/s2
Speed after 1 min V = (accel)(time)
= 011 X 60
= 66 m/s
Example 3. A pin is being driven home by a hammer of mass 1 kg. When
the hammer is moving at 15 m/s it strikes the pin and drives it 10 mm
further in. Assuming the hammer is brought to rest at constant decelera
tion estimate the average force of the blow.
If the hammer travels 10mm (001 m) whilst decelerating at a constant
rate from 15 m/s to rest:
Average speed over the stopping period
= ¥ = 075 m/s
and the space moved = 001 m
Hence:
duration of the blow (t) = ~=j — 00133 s
We then have for the movement of the hammer after it strikes the pin:
initial speed (u) = 15 m/s: final speed (v) =
and time (t) = 00133
and from v = u + at (see p. 10)
= 15 + 00133a
EQUATIONS OF MOTION
from which a = nnn o =  1127 m/s 2 (retardation)
The average force of the blow is found from:
F = (mass)(accel) = 1 x 1127
= 1127N
The equations of motion
The reader, no doubt, will have already realised that the SI system is
more orderly and coherent than the British system and this might be
pursued by considering its application to the equations of motion.
If s = space travelled, v = velocity (or speed) and t = time
then s = vt or v =  (1)
The basic unit for s is the metre (m) and for t, the second. This gives us
g
the secondary basic unit for velocity as v =  = m/s (metres per second).
Naturally, it is not always desirable, or possible, to work in metres and
seconds but alternative units of larger or smaller dimension are avail
able to suit the conditions or aspects of any particular situation, (e.g. for
a road speed we should probably use kilometres and hours) .
Thus: 1 kilometre = 1000 metres
1 hour = 3600 seconds
1 km 1000 m _ 5 m _ 5
1 h ~ 3600 s ~ 18 s ~ 18 m/S
i.e. to convert km/h to m/s multiply by j~.
Acceleration
Acceleration is the rate of change of velocity
, . change in velocity
i.e. acceleration =
time taken for the change
If a body starts from rest and acquires a velocity of v after time t, and
using the symbol a for acceleration
v
a = —
t
KJ
INTRODUCTION
Since the basic derived unit for v is m/s, and the basic unit for t is s, so
the basic unit for acceleration is— — 5 s = — (metres per second squared).
Transposing the above a =  to give v we get v = at.
If, instead of starting from rest, the body already had an initial velocity
of u. Then, final velocity after time t:
f = u + at
(2)
Instead of accelerating a body may be slowing down or decelerating. Then
the acceleration will be a minus quantity and
u — at
This slowing down is termed a retardation.
The above relationships may be illustrated graphically and this is
shown at Fig. 1 .
Fig. 1
Example 4. A drop stamp falls freely for 5 metres under the action of
gravity. Find: (a) the time of fall and (b) its velocity at the instant it
strikes the tup. Takeg = 981 m/s 2 .
ACCELERATION 1 1
The graph representing the fall is shown at Fig. 2 and if v is the final
velocity after time t:
Vel.
O
t
u
\ArM\
5>
s^. 5m\
"
Time
Fig. 2
We get:
1 , C A 10
5 vt = 5 and v = —
t
But for accelerated motion from rest
v = at where a = g = 981 m/s 2
Hence, equating this to (1):
10
= a = 981 t
and
t 2
10
981
Substituting in (1) for t we have:
v —
10
101
t *= 101 s
= 99 m/s
(1)
Example 5. A machine ram is operating on a stroke of 360 mm. It starts
from rest, accelerates at a uniform rate until the centre of the stroke and
then retards at a uniform rate to a standstill at the end of the stroke. If
the stroke occupies 15 seconds find the acceleration and the maximum
speed attained.
12 INTRODUCTION
This problem is best solved with the help of a graph and the motion
is represented in Fig. 3.
Time
Fig. 3
The ram accelerates from to A, its velocity increasing uniformly, and
the reverse process takes place from A to B.
The area OAB represents the space travelled which, in this case, is
360 mm (0.36m).
Hence 036 m = KOB)(AC), and since OB = l5s
036m = (l5s)(AC)
2 x 036 m
l5s
AC = v =
For half the stroke v = at
v 048
Hence a =  = T^^m/s = 064m/ s 2
= 048 m/s
2 Measurement
and gauging
Limits and tolerances
When parts which must fit together are being made under conditions
which do not permit of each fitting pair to be mated up individually, it
becomes necessary to arrange the working dimensions so that if each
component is made to them, the required type of fit will be assured. To
achieve this, a system of limits and fits is adopted. The definitions used
in connection with limit systems will be gathered from the following
and Fig. 4.
, jTolerance
'^///(//s
r V '
__
o
1 1
I 1
oo S
J
1
~i — "^
so J
00 
— i
S9 "^
*«
°> c
os
On .c
o
A ^
 .2>^
^ .o>
fN
ra
"> a:
'.
K>
J
1
** a:
K)
i
'
w/z/Jm
Sha
it 519
n 319
80 t
68 i
1>
Holt
32
! 31 9
15 A
85?
Fig. 4
The low limit is the dimension of the smallest permissible size, and
the high limit the largest permissible size. The component is acceptable
if its size lies anywhere between the two.
The tolerance is the difference between the two limiting dimensions.
[The tolerance on the above shaft is 31980  31968 = 0.012mm.]
The allowance is the variation between the sizes of the hole and shaft
necessary to give the type of fit required. For a running fit the shaft must
be smaller than the hole (clearance fit), whilst for a driving or force fit,
the shaft must be larger than the hole (interference fit). Between the
extremes of clearance and interference there is a range of fits such as
push fit, slide fit, etc., in which there is only a small variation between
the hole and shaft sizes (transition fit).
14
MEASUREMENT AND GAUGING
Limit systems
The numerical value of the limits for any related hole and shaft will
depend on:
(a) The nominal size (i.e. whether 25mm, 50mm, 100mm, etc.)
(b) The class of fit (e,g. running fit, push fit, force fit, etc.)
(c) The grade of workmanship desired.
The ISO system, set out in BS Specification No. 4500 (1969), allows for
27 types of fit and 18 grades of tolerance over a size range of zero to
3150mm. At first sight this seems an enormous provision, but the fits
and grades of workmanship covered allow for everything from fine gauge
work to the roughest form of production, and even for some classes of
raw materials. Average workshop requirements may be met from a
limited part of the specification and for this purpose suitable recom
mendations are given. In the system the 27 possible holes are designated
by capital letters A B C D etc., and the shafts by small letters covering
the same range. The 18 accuracy grades are denoted by the numbers 01,
0, 1, 2, ... 16. The nomenclature adopted for specifying any particular
hole or shaft with its tolerance grade is to write the hole or shaft letter
with its grade number: thus H7 for a hole, or e8 for a shaft of the fit and
accuracy given by the letter and numeral concerned. A fit involving these
two elements is written H7e8 or H7/e8 .
SELECTED FITS (HOLE BASIS)
Type of Fit
Clearance
(Slack,
Running etc.)
Transition
(Push, Slide etc.)
Interference
(force, Driveetc.)
(Details of the limits for the above shafts over a diameter range 6 mm to
250mm are given in Appendix 1, page 225.)
LIMIT SYSTEMS 15
For average workshop use the H hole associated with the accuracy
grades 7 to 11 (H7 to Hll) are recommended as being satisfactory and
details of the limits of these for a diameter range of 6mm to 250mm are
shown in the table on p. 226.
140
120
100
80
60
40
20
E °
E
S 20
o
£ 60
v.
®
"o 80
100
120
140
160
180
200
220
240 
ill
Clearance
Transition
Interfere/! ce
Fig. 5 ISO System of Limits and Fits. Hole and shaft relationships for selected fits (hole
basis). (Tolerance scale applies to the diameter range: 18mm— 30mm.)
16
MEASUREMENT AND GAUGING
From the shafts included in the specification the table gives a selec
tion of those recommended as likely to be the most useful in coping with
the needs of the average workshop. The holes with which they should be
associated as well as the approximate type of fit are also given.
Naturally from a selection of 27 fits associated with 18 grades of ac
curacy it is possible to choose many combinations, but the small selec
tion above should be sufficient to help the reader in his study of the
subject. Fig. 5 illustrates the hole and shaft relationship for a selection
of the fits associated with the H7 to Hll holes as recommended in the
BS Specification. From the diagram the reader will be able to trace the
maximum and minimum metal conditions for the range of diameters to
which the details apply.
Exercises 2a
Write down for problems Nos. 1 to 4 the hole and shaft limits, and calculate the maxi
mum and minimum clearance or interference. (See tables in appendices)
1. 75mm BSI H10 hole and e8 shaft.
2. 35 mm BSI H9 hole and k6 shaft.
3. 20mm BSI H7 hole and f7 shaft.
4. 57 mm BSI H8 hole and s7 shaft.
5. Two 32mm BSI H8 holes are bored in a plate at 105 mm ±002 mm centres. Their
centre distance is to be checked by a gauge of the type shown at Fig. 6. If the "go" ends
Fig. 6
of the checking plugs are 12 mm diameter calculate their "Not go" diameters and their
centre distance.
6. To allow for gauge wear the "Go" end of a limit plug gauge, made for a 40 mm BSI H9
hole, is made larger than the minimum hole size by 10% of the tolerance. Calculate the
diameter of this end of the gauge.
7. A limit caliper gauge made for a 50 mm BSI f7 shaft is to be ground out to suit a
k7 shaft. Determine the alteration necessary.
8. Two blocks each 25 mm wide are made to BSI h7 limits and a 50 mm slot is made
to H8 limits. Compare the fit of these blocks put together with that of a single 50 mm block
SLIP GAUGES 17
made to h7 limits. (Assume the width of the blocks put together to be equal to the sum
of their individual thicknesses.)
9. A number of limit plug gauges are available, made for the former Newall fin.
Class B hole ( — in). Could these be ground down to suit a 19mm BSI H8 hole
and, if so, what alteration would be necessary?
10. A BS 40mm b9 shaft is placed in an H8 hole. Determine the greatest and least
clearance possible between the shaft and the sides of the hole.
Slip gauges
For the purpose of checking the accuracy of micrometers, verniers and
other gauges, it is necessary to have available some means of building
up any required length. In most workshops this is achieved by the use of
slip gauges. These consist of blocks of different thicknesses, which
are made to such a fine degree of accuracy and flatness on their measuring
faces, that they may be "wrung" together and the overall length of any
number of blocks so joined is the sum of their individual lengths. These
gauges are often called Johannsen gauges after their originator. (Fig. 7)
The Coventry Gauge and Tool Co. Ltd.
Fig. 7 A set of slip gauges (107 pieces).
{Lid of box not shown)
The number of blocks in a set will depend upon the range of sizes that
is required to be made up, and a mediumsized set of such gauges contains
47 pieces of the following sizes.
18 MEASUREMENT AND GAUGING
Metric dimensions: widths of bloeks in mm.
Pieces
Range
Steps
1
1005
—
9
101109
001
9
1=119
01
24
124
10
4
25
, 50, 75, 100
250
47 Total
From such a selection it is possible to choose blocks to build up almost
any dimension that can be named within the capacity of the set. It will
be seen that there will be alternative methods of making up a size, but
the one should be chosen which employs the smallest number of pieces.
Example 3. Choose blocks to assemble the following sizes: (a) 37.31 mm
0)6 1685 mm.
To assemble (a) we may use
the following blocks:
101
13
1000
2500
3731
Dimension
(b) may
be obtained
as follows:
1005
108
16
800
5000
61685
The above example should indicate to the reader how to proceed for
any other size, the method being to start at the figure on the extreme right
of the dimension required, choosing gauges to accommodate each
figure in turn.
English sizes
For routine English work, slip gauges made to inch dimensions may be
obtained but it is possible, by calculating the metric equivalent, to use
gauges from a metric set.
Example 4. Make up a set of slip gauges to check a gauge measuring
2 1758 in.
THE SPIRIT LEVEL
19
The metric equivalent of 22758 in is found by multiplying 21758 by
254 and is equal to 5526532 mm.
The nearest size to this, that may be assembled from the set of gauges
given above, is 55265 which is 000032 mm or 0000012in. too small.
Gauges to make up to 55265 are as follows:
1005
106
12
2
50
55265
Exercises 2b
From the list given on p. 18 make up sets of slip gauges to give the following sizes:
1,1111 mm 2. 2364 mm 3. 35635 mm
4. 6878 mm 5. 7570 mm 6. 11 536 mm
7. 9.52mm 8. 44.45mm 9. 80.99mm
34925
10. A dimension given as , 4 q mm is to be checked. Make up two separate sets of
blocks, one to measure each limit.
11. Make up sets of slip gauges to check the jaws of a limit gap gauge for a 30mm BSI
f7 shaft.
12. Two holes, 0875 inch diameter and 10625 inch diameter, are bored in the face of a
casting at 15625 inch centres. Convert to mm, and make up a set of slip gauges to test
between the insides of test plugs placed in the bores.
The spirit level
The spirit level consists essentially of a glass vial fixed into a frame. The
inside top surface of the vial is not straight but is formed to a radius, con
vex upwards, as shown in Fig. 8. In some instruments the inside of the vial
is ground barrelshaped as shown, whilst in others the glass tube which
Fig. 8
20
MEASUREMENT AND GAUGING
forms the vial is bent to a radius. The vial contains spirit with sufficient air
space to leave a bubble, and is cemented into the frame and accurately
located so that when the base of the frame is level the bubble rests at the
centre of the scale.
The relations governing the movement of the bubble and the angle of
tilt of the level will be followed from Fig. 9 and the following:
fait««t
Bj_J
""77777777777777777777777777^77777777777^1^
Fig. 9
Arc CAD represents the upper inner surface of the vial and O the centre
of its radius. OB represents the base of the level. OB is horizontal and
perpendicular to OA. If OB in now tilted to bring B to B,, point A on the
vial will swing to A, but the bubble will remain vertically above O and will
travel along the vial to A.
A,A
If is the angle of tilt, then 6 (radian) =
R
If BB, is the length of the arc through which one end of the level (length
BB,
L) swings, then 9 (radian) = — jr±
BB,
L
AA,
R
and BB, =
L. AA,
R
Actually, the height h that one end of the level is above the other is the
dimension we require, but when dealing with angles as small as those con
THE SPIRIT LEVEL 21
cerned here, the difference between BBj and h is so small as to be
negligible.
Thus we can say that h = ^(distance bubble moves)
If the angle of tilt is required in degrees, then since
Movement of bubble
(radian) =
9 (degree) =
R
573 (Movement of bubble)
R
Unfortunately, not many makers of levels mark them with particulars as
to the radius (R) of the vial, but this may be determined experimentally by
tilting one end of the level by a known amount and after noting the move
ment of the bubble, calculating the radius from the above expressions.
Example 5. A spirit level is 300 mm long, and it is found that when one end
is raised 002 mm above the other, the bubble moves 150 mm along the
vial. Calculate the radius of the vial.
, 17 u . , . h Movement of bubble
We have that y = =
002 _ 15
300 ~ R
From which R = ^p^' 5 = 22 500mm or 22 5m
Example 6. The base of level is 450 mm long and the radius of the vial is
30 m. Find (a) the height of one end above the other, and (b) the angle of
tilt, corresponding to a bubble movement of 3 mm.
h _ Movement of bubble
T = R
* , _ L (movement of bubble)
R
450 x 3 ....
0 045mm
30 x 1000
Angle of tilt in degree = 3 q ? ^ ^qq^ = 0005 73° = 206 second
22 MEASUREMENT AND GAUGING
The sine bar
For accurate work in connection with angles the sine bar possesses
advantages over the usual forms of protractor. Sine bars differ in form,
but the considerations affecting their setting are the same in every case.
Two common types of sine bar are shown in Fig. 10.
The bar shown at (a) has two plugs which are let in and project about
12 mm from the front face. At (b) is shown a bar which is stepped at the
ends and a roller is secured into each step, being pulled in by a screw so as
to contact with each of the faces of the step. Both at (a) and (b) the follow
ing points are important if the sine bar is to be of any use:
(i) The rollers or plugs should both be of the same diameter.
(ii) Their centre distance must be absolutely correct. (The diagram is
dimensioned as 200 mm centres, but sine bars are available in 100, 250 and
300 mm centres as well.
(iii) The centre line AB of the plugs must be absolutely parallel with
the edge of the bar used for measuring (generally the bottom). It is desir
able for the two edges of the bar to be parallel, with AB parallel with
both.
Fig. 10.
When in use, the bar shown at (a) lends itself to clamping against an
angle plate, whilst that at (b) can be rested on two piles of Johannson
gauges to give it the correct inclination.
Calculation for sine bar setting
In Fig. 11, C is the centre distance of the plugs, h is the height of one plug
above the other and a is the angle to which the bar must be set.
CALCULATION FOR SINE BAR SETTING 23
Fig. 11
Then
QR h
and h = C sin a
i.e. difference in height of plugs = (centre distance) (sine of angle).
Example 7. Calculate the setting of a 200 mm sine bar to measure an angle
of 36° 38'.
We have tljat sin 36° 38' = 05967
h = 200 X 05967 = 119.34
Hence one plug mu§t be set 11 934 mm above the other.
Example 8. Calculate the setting of a 250 mm sine bar to check the angle
of g taper of 1 in 16, ^n tjie diameter.
The taper is shown in pig. 12(a) and, if it is assumed to pe 16 units long,
then in triangle ABC,
16
Fig. 12(a)
AC = 16, BC = 4, and BAC =
ta 44 = 3i =o  03125
24 MEASUREMENT AND GAUGING
From which  = 1° 47' 30"
and 6 = 3° 35'
Now sin = sin 3° 35' = 00625
and since the setting is for a 250 mm sine bar:
h = 250 X 00625 = 15 625 mm
Fig. 12(b)
The setup shown at Fig. 12(b), A and B being two sets of block gauges
assembled to give the setting calculated above.
Precaution when checking plane surfaces with the sine bar
The reader should observe that when using the sine bar to check the angle
between two plane surfaces (e.g. the surfaces of an angle plate) the bar
must be set accurately at right angles to the slope of the face being
measured. The following example will illustrate how an error may be in
troduced if this is not done.
Example 9. A surface was being checked by a 100 mm sine bar which, due
to an error in setting, was placed 6 mm out of square with the slope of the
surface. The angle obtained from the sine bar readings was 59° 30'. Find
the true angle of the surface being measured.
The conditions are shown diagrammatically, and exaggerated, in Fig.
13.
The readings should have been taken on the line of greatest slope AB,
but were actually taken on AC.
CH is a perpendicular drawn from C on to AB, and from the conditions
of the problem, CH = 6 mm.
AB == AC = 100 mm, E, F and G are points where horizontal lines
through A meet verticals through B, C and H.
GAUGING LARGE RADII 25
Fig. 13
In triangle ACH: AH 2 = AC 2  CH 2
100 2  6 2 = 9964
AH = \/9964 = 9982
Now since CH is parallel to the slope, C and H are the same height and
CE = HG
In triangle CAE:
AC = 100, and CAE = 59° 30'
.'.CE = 100 sin 59° 30'
= 100 x 08616 = 8616mm
Then since CE = HG; in triangle HAG
HA = 99 82 mm, HG = 86 16 mm and HAG is the true angle of the plate
Sine HAG = QQ _. = 08631 from which, HAG, the true angle of the plate
= 59° 40'
i.e. an error of 10'
Gauging large radii
From some classes of work it is necessary to measure the radii of circles
which are too large to be straddled by calipers or a micrometer. Where the
complete circle is available its circumference may be measured by a tape
and the diameter obtained by dividing by n. This method, however, is by
no means perfect, and is not at all convenient when the radius to be
measured is not part of a complete circle.
An alternative method is to determine the radius by reference to the
distance of its surface from the corner of a vee block resting on it.
26
MEASUREMENT AND GAUGING
In Fig. 14 ABC represents the faces of a vee block resting on a circle,
centre O and radius R.
r^'i
Fig. 14
If 6 is the included angle of ABC, its half angle ABO will be ■=• as shown.
The circle and the block contact at D, and in triangle BOD, since D is a
right angle, angle BOD = 90  ^
For differentsized circles placed in the vee, the variable length that
we can measure is BE, so that the problem becomes one of finding a suit
able expression for R in terms of this length and the angle of the vee.
Now BE = BO  EO and EO = OD = R
BO
But
OD = secant ( 90  l)
.\BO=ODsec (90  )
= R cosec y
Npte:
secant =
cos
since sec ( 90 
!)
cosec 7T
Hence
BE = i?cosecy
R
cosecant = r—
sin
iacosecy— lj
From which
R =
BE
cosec
GAUGING LARGE RADII
27
For any given angle of vee, the quantity cosec ^  1 is constant and
can be calculated and stamped on the gauge. All that is then necessary is
to measure BE, and divide it by this number to give R.
Example 10. A gauge of the type shown in Fig. 14 having an included angle
of 120° is placed on a tube and the length BE measures 2 125 mm. Find the
diameter of the tube.
Here
R =
BE
2125
2125
cosec y — 1
cosec 60°  1 11547  1
2125
01547
= 13736 mm
and
dia. of tube = 2 x 13736 = 27472 mm
The main difficulty in the use of a gauge of this type is the accurate
measurement of the distance BE. This may be overcome by constructing
the gauge and incorporating a depth gauge or micrometer head on the
centre line at B. An alternative construction is to make the gauge with a
fiat portion as shown at Fig. 15, using slip gauges to check the distance
between the work and the flat.
It will be an interesting example to plan such a gauge as this.
Example 1 1 . Plan out a vee gauge to measure round work, to cover a
range of diameters varying from 250mm to 750mm.
Fig. 15
28 MEASUREMENT AND GAUGING
We will make the gauge with an included angle of 120°, and the flat
portion FG filling in the corner of the vee such that it just contacts with
a 250 mm circle placed in the vee, as shown in Fig. 15 . The length of the
gauge must be sufficient to accomodate a 750mm circle.
The length must exceed twice the distance from E to line AB, i.e. > 2AE
sin 30° = > 2 x 375 x h say, 400mm.
If flat FG just touches a 250 mm circle placed in the vee,
then
CB = CD sec 30° = 125 x 11547 = 144.34mm
and from the vee corner B to FG the distance is 14434  125 = 1934 mm.
It now remains to find the relation between the radius of any circle
placed in the vee, and the distance from its circumference to line FG.
From Fig. 14 we have that from B to the circumference of a circle radius
R,
= R cosec j — 1
a
which becomes 01547/? when j = 60°.
We have to subtract 1934 mm from this, so that
d = 01547/?  1934 mm
and transposing to give R results in
D d+ 1934
R = 01547 mm
The distance d can be measured with slip gauges, or a micrometer depth
gauge incorporated in the construction as before. By marking the above
formula on the gauge, the checking of sizes becomes a routine job.
The measurement of large bores
An interesting example of the measurement of a large diameter is avail
able to us in the gauging of large holes with a point gauge . This is shown in
Fig. 16(a), where a hole of diameter D is being gauged by a point gauge
of length L. In practice, L is a very small amount less than D, and when
the gauge is held at one end, a small amount of rocking movement on
either side of the centre line is possible at the other end. This is indicated
byw.
THE MEASUREMENT OF LARGE BORES 29
Fig. 16(a)
The conditions are shown exaggerated at Fig . 1 6(b) . The full circle is the
hole being gauged, and the dotted circle (centre A) is that which the end of
the point gauge would describe if it made a full sweep . Actually, the end of
the gauge only moves over the arc BHF.The amount by which this guage
is smaller than D is shown by CH. Let this be S.
Fig. 16(6)
An approximate solution to the problem which will be accurate enough
for most practical purposes is as follows:
30 MEASUREMENT AND GAUGING
If BC is joined, angle B is a right angle, since it is the angle in a semi
circle, and in triangle ABC: AC 2 = CB 2 + AB 2 .
AC = D; AB = L and CB is very nearly equal to w.
Hence we may say: D 2 = L 2 + w 2 approximately.
But D = L + 8
(L + S) 2 = L 2 + w 2
L 2 + 2LS + S 2 = L 2 + w 2
Now S will be a very small quantity, probably less than 002 mm so that
S 2 will be so small that we may ignore it.
Hence U + 2LS = L 2 + w 2
2LS = w 2
S = 2L
This enables us to find the amount the gauge is smaller than the hole
when we know how far the end may be rocked on either side of the centre
line.
Example 12. If a 375 mm point gauge rocked 6 mm at one end, calculate
the diameter of the hole being gauged.
w 2
The difference (S) = ^ and if the total movement is 6 mm, w — 3 mm
3 2 9
' ~ 0012mm
* 2 X 375 750
Hence, Hole diameter = 375012mm
Exercises 2c
1. The radius of the vial of a spirit level is 25 m. When this is placed on the bed of
a machine 2^m long, the bubble is 3mm from its central position. What is the end to
end error in the machine?
2. A machine bed is 18 m long and is tested by a level 150mm long. One division on
the level corresponds to an inclination of 006 mm per m. The level is transversed in
steps of its length, from the LH to the RH end of the bed, with the following results:
Position of Level
1
2
3
4
5
6
Reading
(Division)
+ LH end high
 RH end high
+ k
+ 1
+ 1*
+ i
(continued on p. 31)
THE MEASUREMENT OF ANGLES AND LARGE RADII 31
Position of Level
Reading
(Division)
+ LH end high
 RH end high
1
10
U
11
1
12
Make a scale diagram showing the dip in the bed to an enlarged scale, and calculate
the total error.
3. If a level is to be sensitive enough to indicate 1 minute of angle by 3 mm of move
ment of the bubble, what must be the radius of the vial?
4. Calculate the setting of a 250mm sine bar to check a taper of 1 in 6 on the
diameter.
5. A 100 mm sine bar is used to check the inclination of a surface which is given on the
drawing as 26°36' ± 4'. The height of one plug above the other is found to be 4503 mm.
What is the error in the angle of the surface?
6. A sine square has 4 plugs spaced at the corners of a 125 mm square. To mark out the
template shown at Fig. 17, the template is secured to the sine square and line AB is
first marked parallel to two of the plugs 1 . The square is then tilted and set to the correct
angles for marking BC, CD and AD. Calculate the settings for marking these lines and
also the lengths x and v.
Fig. 17
7. In testing an angular surface with a 200 mm sine bar, the reading obtained is 12230 mm
for the height of one plug above the other. If the bar is 6 mm out of alignment with the line
of maximum slope, calculate the true angle of the surface and state the error in the
reading.
8. A point gauge 500 mm long, when tried in a bore, rocks a total amount of 12 mm
at one end. When it is tried in a position at 90° to the first position, the movement is
24 mm. Calculate the mean diameter of the bore and the out of roundness.
9. A vee of 120° with a 12mm flat at its bottom is placed on a cylinder and the distance
from the flat to the curved surface of the cylinder is 240 mm. Calculate the diameter of
the cylinder.
32
MEASUREMENT AND GAUGING
10. In Fig. 18 A and B are two spherical seating pins 14 mm diameter.
Calculate the height H so that the distance between the circle and the setting pin C
shall be 050 mm.
(A and B are true half spheres.)
50
Fig. 18
11. How much rock must be allowed on a point gauge 460 mm long if the bore is to
be finished to 46006 mm diameter?
The location of points on angular surfaces
When a component having an angular surface is shown in orthographic
projection on a drawing, only the projected view is given, and unless a
true auxiliary view is added to give particulars of points located on
the surface, some means must be found to calculate their position.
Two or three problems of this type will be illustrated in the following
examples:
Example 13. The plan and elevation of a block with two holes is shown
in Fig. 19 (a).
Find (a) the centre distance of the holes,
(b) the angle between XX and AB, both dimensions as measured
on the sloping surface.
In dealing with problems of this type it is well to cultivate the sense of
visualizing lengths in 3 perpendicular directions, and also of being able
to make a rough pictorial sketch of the data. The sketch is generally
very useful in helping to show up how the problem should be treated.
In Fig. 19 (b), A and B are the hole centres, C is a point on the sloping
surface where the centre lines intersect, and D is the point where a vertical
through A meets a horizontal through C.
THE LOCATION OF POINTS ON ANGULAR SURFACES 33
 V ?
(a)
Fig. 19
(b)
Then in triangle ACD:
A A
CD = 26 mm, D = 90° and C = 35°
AC
CD
26
cos 35° 08192
= 3174mm
The centre distance AB is the hypotenuse of the rightangled triangle
ABC.
AB 2 = 26 2 + 3174 2 = 1684
AB = n/1684 = 4104 mm
The angle made by AB with the side of the block is the angle ABE and
AE • ABC 26
AB = Sm ABE = 4T04
= 06335
/\
From which ABE = 39°18 /
Example 14. Fig. 20 (a) shows the plan and elevation of a block in which
a hole has to be drilled, entering at the point A, and leaving at B.
Calculate the angular settings of the block for drilling.
34
MEASUREMENT AND GAUGING
Fig. 20(a)
Fig. 20(b)
The centre line of the hole is lettered XY in the elevation, and X,Y, in
the plan, point B being underneath the block as drawn. A pictorial view
of the job is shown at Fig. 20 (b), and in that diagram AC is a vertical
and BC a horizontal line. For drilling the hole, the base of the block must
be set at an angle of 9, and the face containing A must be set at a, both
angles being with the vertical.
In triangle CDB:
A
CD = 29 mm, DB = 19 mm and D = 90°
CB 2 = 29 2 + 19 2 = 1202
CB = ^1202 = 3467 mm
tan =
AC 1 15
CB 3467
18°21'
03317
DB 19 ft ,_„
tan a = — = 29 = ° 6552
a = 33° 14'
Example 15. Calculate the angular settings for drilling a hole on the line
AB shown in Fig. 21 .
If A is joined to C, triangle ABC is formed and AC is parallel and
equal to the line DE, shown on the end of the bar.
THE LOCATION OF POINTS ON ANGULAR SURFACES 35
DE = OEsec 45° = OE x 1414 = 26 x 1414 = 3676 mm
and in triangle ABC:
BC + A 52 , „,„,
AC = tanA = 36^76 =M146
from which A = 54° 45'.
Hence, if lines DE and EC are marked on the bar, and the end of the
bar set at 54° 45' to the horizontal with DE parallel to the line of greatest
slope, a hole started at B will break through at A.
Fig. 21
Example 16. In the drawing of the component shown at Fig. 22 (x)the
two sloping holes starting at A and B must meet at a point 10 mm from
the base of the block. Find the starting heights a and b of the angular
holes.
A diagrammatic view of the base of the block is shown at Fig. 22 (y)
in which the lines AB, BC and AC are assumed to be horizontal ones (i.e.
the projections of lines joining the points A, B and C).
36 MEASUREMENT AND GAUGING
Fig. 22
In triangle ABD:
AB 2 = 13 2 + 16 2  2 x 13 x 16 cos 60° (cosine rule)
= 169 + 256  208 = 217
AB = v^217 = 1473 mm
also
AB
13
sin 60° sin B
. a 13 sin 60° ._,..
sin B = — rjss — = 07644
1473
B = 49°51'
A
Hence A = 180°  (60° + 49° 51') = 70°9'
/\ /\
Since DBC = 90°; ABC = 90°  49°51 / = 40°9'
/\
also BAC = 90°  70°9'  19°51'
In triangle ABC; AB = 1473. B = 40°9' A = 19°51'
THE LOCATION OF POINTS ON ANGULAR SURFACES 37
also C = 180°  (19°51' + 40°9') = 120°
BC 1473
and  — irtog1 , = : — r^rr from which BC = 577 mm
sin 19°51' sin 120°
also in the same triangle:
AC 1473
sin 40°9' sin 120°
whence AC = 1097 mm
The centre lines of the holes of which AC and BC are projections, slope
upwards at 20° and 30° respectively.
Hence, height of A above C = AC tan 20° = 1097 x 0364 = 399 mm
and height of B above C = BC tan 30 = 577 x 05774 = 333 mm.
This completes the solution of the problem and gives us the following
data:
dimension a = 399 + 10 = 1399 mm
dimension b = 333 + 10 = 1 333 mm
Exercises 2d
1. A straightedge is placed on a surface sloping at 36° 45' and is set at an angle of
15° to the line of greatest slope. Calculate the inclination of the straightedge to the
horizontal.
2. In Fig. 23 a hole is to be drilled, starting in the centre of the sloping face, and
breaking out at the corner B. Calculate the angle between the vertical plane containing
centre line AB of the hole, and the face BCD of the block.
^*^*^ D_^t^>^
,/
civ?
a V /
( l~
51 Jf
Fig. 23
3. For Fig. 23, calculate the angle between the centre line of the hole and the base
of the block, for a hole starting at A and breaking out in the centre of the base.
38 MEASUREMENT AND GAUGING
4. In Fig. 24 A and B are two points on the sloping surface shown, and line AB is
parallel to the sloping edge. Find (i) angle line AB makes with the horizontal, (ii) dimen
sions a and b, and (iii) distance AB.
Fig. 24
5. In Fig. 25, 6 holes are equally spaced, start on an 80 mm pitch circle and break
through on a 130 mm circle. Calculate the angle between the centre lines of one hole
and the next.
6. In Fig. 26 find the distance a and the angle a for a hole whose centre line AB shall
be tangential to the 60mm circle.
7. Calculate the height from the base to the centres of holes A and B in Fig.
27.
8. In Fig. 28 a hole AB starts at A and leaves the bottom of the block at B. Find the
length of this hole and the angle it makes with the base of the block. Another hole is to
start at H, and be drilled in a plane parallel to the plane CDEF. The second hole must
run into the first (i.e. their centre lines must intersect). Calculate the angle that the centre
line of the second hole must make with the base.
Measurement of tapers by means of balls and rollers
Male taper with rollers (Fig. 29 (a)). If two similar rollers are placed in
contact with the taper on opposite sides as shown, then for rollers of
diameter d and centers c:
h = c tan
(1)
MEASUREMENT OF TAPERS BY MEANS OF BALLS AND ROLLERS 39
Fig. 25
Fig. 26
4 Holes
equally
spaced
165 f P.C.Crs.
36"
105
Vertical height
from base
Fig. 27
Fig. 28
40 MEASUREMENT AND GAUGING
and the difference between the dimensions taken over the top and bottom
pairs of rollers will be 2h. When the taper is dimensioned as 1 in a certain
length (say 1 in /) on the diameter,
then
 = 2/ and A = 27
(2)
Fig. 29 {a).
A taper given in mm/ unit length may be converted to 1 in / by divi
sion.
In some cases it may be necessary to have a check on the diameter D
of the taper, and if this diameter is situated at a distance H from the centre
of the top pair of rollers:
In triangle BEF:
BF a
BE = COS 2'
and since BF = A dia of roller = 
2
yop = cos y, from which BE
2 cosy
The radius EL of the taper = .  GK (EG is parallel to the taper
centre line).
T , T D _„, a D .,, a
EL = _  EG tan ■_ = =  H tan y
MEASUREMENT OF TAPERS BY MEANS OF BALLS AND ROLLERS 41
D.
C (centre distance of rollers) = 2 (BE + EL)
V 9 r»r»c
In practice, C, d and H would be known, so we require to transpose for
(d D „, a\
_ + __// tan y \
T /
(3)
This gives that
and
C „ A a d D
=■ + H tan T = =
2 2 „ a 2
2 cosy
D = C+ 2#tan£ —
2 or
cos T
(4)
If the taper is dimensioned as 1 in /:
Then
z
/
a 1
tan 2 = 27
and
D then becomes
a
COS^r =
2 vTTT 2
(Fig. 29(A)).
H /fv/T+72
D  C + —  *
(5)
Fig. 29 (6).
In practice, for the measurement of tapers in this way, it is helpful to have
a fixture of some kind which will support the taper and provide supporting
and gauging arrangements for the rollers.
42
MEASUREMENT AND GAUGING
Example 17. A taper of 1 in 10 on the diameter is 80 mm long. Two pairs
of rollers 12 mm dia. are used to check it, and the spacing of the rollers is
as shown in Fig. 30. If the reading over the top rollers is 105 mm, cal
culate (a) the reading for the bottom rollers, (b) the diameter of the taper
on a circle 14 mm from the top.
Fig. 30.
If the top rollers are situated at 58 mm from the face on which the
taper and bottom rollers are resting, the roller centres will be 52 mm as
shown .
Centre distance of top rollers = 105  12 = 93 mm and applying the
formula to find h we have
h = ^[where c = 52 and / = 10]
52 i *n
= ~?j = 260 mm
Hence the reading over the bottom roller will be
93  52 + 12 = 998 mm
For calculating the diameter D of AB we have from (5) above that
H dV L 4 I 2
and C  93, H = 8, d = 12, / = 10
CHECKING A TAPER HOLE BY MEANS OF BALLS 43
So that
8 12VT+100
D ~ 93 + To 10
= 938  12012 = 81 788 mm
Checking a taper hole by means of balls
For this it is necessary to use two balls of different diameters which will
rest in the hole, touching its sides. The ball sizes should be chosen to give
a reasonable centre distance (c), and this may be measured by employing
a depth gauge from the top face of the hole to the top of the lower bah\
and a height or depth gauge to the top of the upper one.
Fig. 31
Generally, the dimensions R, r, c and h in Fig. 31 would be known, and
we require to derive formulae for finding a and D.
E is the point where the ball contacts with the side of the hole and BC
is parallel to the side of the taper.
Then in triangle ABC:
AB = c y C = 90°, and B
so that
AC
AC
AB
R — r (since CE = r),
R
. a
sin j
— r .a
= Sin fr
c 2
(6)
which enables us to find the angle a of the taper.
44 MEASUREMENT AND GAUGING
If the taper is dimensioned as 1 in / on the diameter, then from
Fig. 29 (b),
. a i
sin^ =
and from above,
2 ^TTT 2
R  r i
vTTT 2
vJTT 2 =
c
2(R  rY
Square both sides:
4{R  r) 2 '
from which / 2 =
c*
(7)
4(i?  r) 2 4
and / = ,/ c2  J c2 ~( R  V
 Vc2 ~ ( R ~ r ) 2
2{R  r) ■
To obtain an expression for the top diameter D of the hole we must
consider triangles AEF and FGH
and D == 2(AF + GH)
= 2^AE sec j + FG tan y J [since E = G = 90°]
= 26? sec y + Atan f) (8)
For a taper of 1 in / on the diameter
sec j = * + F and tan  = 1 [Fig. 29 (6)]
and D becomes 2 p ?V ' + /2 + Al
(9)
Example 18. In a check on a taper hole, using the symbols and method
given in Fig. 31, the following results were obtained:
CHECKING A TAPER HOLE BY MEANS OF BALLS 45
R = 15 mm, r = 125 mm, c = 374 mm and h = 73 mm.
Calculate the total angle of taper and the top diameter.
Here we have from above, that sin y =
15  125
~ 374
This gives y = 3° 50' and a = 7° 40'.
= 00668
D = 2(rscc^ + /itan^J
sec^= 10022
tan j = 00670
D = 2(15 x 10022 + 73 x 00670) = 3 104 mm
Exercises 2e
1. Calculate the diameter D and the included angle of the taper for the case shown in
Fig. 32.
Fig. 32
Fig. 33
2. For the gauge shown in Fig. 33 calculate the centre height H between the two balls.
(See Example 22, p. 52.)
46
MEASUREMENT AND GAUGING
3. Calculate, for the setup shown in Fig. 34, the dimensions A and B. (The rollers
are all 20 mm dia.)
4. For the example shown in Fig. 35, make up an expression connecting H with
the centre distance D of the rollers. Find H when D = 77 mm.
Total. Taper
1 in 20
'777777777777?77777777)(7777777777777777777
58 <p A
Fig. 34
Two 20 Rollers
Fig. 35
5. A hole A of diameter D has two equal plugs, B, which just fit into it. Obtain
an expression in terms of D for the diameter d, of two more plugs C, which will just
fit into the spaces left (Fig. 36).
Find d when D = 80 mm.
Fig. 36
Fig. 37
6. In Fig. 37 the three radii blend together.
Calculate the angle a.
WIRE MEASUREMENT OF SCREW THREADS 47
Wire measurement of screw threads
For the fundamental accurate measurement of a screw thread such as
is necessary in turning taps and screw gauges, the method employing 3
wires is a very useful one. The wires are arranged as shown in Fig. 38 (a).
We will work out a general case, and in Fig. 38 (b) is shown a wire,
radius r, resting in a sharp pointed thread of angle a, pitch/? and effective
(or mean) diameter D E . The wire size is not important, providing the
three are the same diameter, touch at the flat portion of the thread, and
are large enough to project above the thread for gauging. The best results
(a)
Fig. 38
(*)
are obtained, however, if the wires touch the thread at the effective
diameter, and for this reason the wires should be near to the following
diameters:
For ISO Metric and Unified d = 0577 p
For Whitworth d = 0564 p
In Fig. 38 (b), AD
AB cosec y = r cosec ■=
H
DE cot y = f cot 7
CD = \H = 4 cot
h = AD  CD = r cosec j  j cot x
48 MEASUREMENT AND GAUGING
and distance over wires (W)
= D E + 2h + 2r
= d e + 2 ( r cosec f ~ f cot y) + 2r
= D E + 2r(\ +cosec^cot
= D E + d\l + cosec 2)~J cot f ( wnere d = dia. of wires)
Having established this general formula, which may be applied to any
thread, we will determine its special adaptation for the most common
thread forms.
(a) ISO metric and unified (Fig. 39)
Here D E = D  2(0325/>) = D  065p
a = 60°
cosec y = 2 : coty = 1732
W{over wires)
2 COt 2
= Z) £ + d (1 + cosec y J
= D 065p + 43)  (1732)
= D + 3d  1516/j
<
J
6l
6
'^fr^Vv
1
'
p
Fig. 39 ISO metric and unified
Fig. 40 Whitworth
CHECKING THE THREAD ANGLE OF A SCREW 49
(b) Whitworth (Fig. 40).
Here the depth of thread is 064/7, so that D E = D  064/?
Also since a = 55°
cosecy = 21657
cot ^ = 1921
fF(over wires) = D E + d(l + cosec y] — y cot y
=  064/? + </(31657)  L 1921
= Z) + 31657rf  l60/>
Example 19. Determine the measurement over wires for the following
cases:
(a) Af30 x 35 ISO metric using wires 2 mm dia.
(b) 1 in dia. X 10 t.p.i. Whitworth using wires 0062 in dia.
(a) M30 x 3.5 ISO metric using 2mm wires.
W = D + 3d  1516/?
D = 30, d= 2 and/) = 35
.. W= 30+ 3 x 2  1516 X 35
= 30694 mm
(b) 1 in dia. x 10 t.p.i. Whitworth using 0062 wires
W = D + 31657rf 160/7
D = \,d = 0062 and/) = 01
W = 1 + 31657 x 0062  160 x 01
= 1036 in.
Checking the thread angle of a screw
By taking measurements over two sets of wires of different diameters
a check on the thread angle may be made. The underlying theory of this
is similar to that given on page 43 for the measurement of taper holes
by means of two balls.
50
MEASUREMENT AND GAUGING
Miscellaneous problems in measurement
Examples in gauging and measurement are so many and varied that it is
impossible to establish set rules for application to every problem. If the
fundamental rules of geometry and trigonometry are known thoroughly
it is nearly always possible to apply some of them to the solution of the
problem. The following miscellaneous examples will serve to indicate
possible methods of dealing with problems of a similar type.
Example 20. Two 25 mm circles centres D and B, touch a 100 mm circle,
and their centres subtend an angle of 50° at its centre. Find the diameter of
a circle which will touch the other three. The problem is shown at Fig. 41
and we require to obtain the diameter of circle C .
Fig. 41
If C and B are joined and triangle ABC is considered, we have:
AC = 50 + r; CB = 125 + r, AB = 625 and A = 25° (r = radofC)
Applying the cosine rule for triangle ABC
CB 2 = AC 2 + AB 2  2AC.AB cos A"
Substituting the values from above gives us
(125 + a) 2 = (50 + r) 2 + (625) 2  2(50 + r)(62.5)(09063)
MISCELLANEOUS PROBLEMS IN MEASUREMENT
51
Multiplying out the brackets gives:
1563 + 25r + r 2 = 2500 + lOOr + r 2 + 3906  5664  11329r
eliminate r 2 from both sides and collect terms
25r  lOOr + 1 1329 r = 2500 + 3906  5664  1563
3829r = 5857
5857
r =
3829
= 1529 mm
giving a circle of 2 x 1529 = 3058mmdia
Example 21. From a piece of round material 40 mm radius, a piece 50 mm
radius is cut as shown in Fig. 42. Find the distance x .
Fig. 42
If AC and BC are joined as shown in the figure, then in triangle ABC:
AB = 59, AC = 40 and BC = 50
also AC 2 = AB 2 + BC 2  2AB.BC cos B (cosine rule)
D AB 2 + BC 2 AC 2 59 2 + 5 2 4 2 ,. . ,
cos B = „ . ^ = — — — (working in cm)
2 AB.BC
4381
2x59x5
59
A
B = 42°3'
If CD is 1 r to AB, then
= 07425
DB = CB cos B = 50 cos 42°3'
= 50 x 07425 = 37 125 mm
x = AB  DB = 59  37125 = 21 875 mm
52
MEASUREMENT AND GAUGING
Example 22. A profile gauge is as shown by ABCD in Fig. 43. Two plugs
are placed in the gauge and dimension h is required as a check.
Fig. 43
In the diagram, EF and HM are drawn parallel to AB and PG in per
pendicular to it. HN is a horizonatl line, and if we can find GN we shall
be able to determine h.
CK
26 x tan 32° = 1625 mm
In triangle JCG: JCG = ^ = 61°, JG = 125 and JC = JG cot 61° =
125 cot 61° = 693mm
In triangle ALB: LB = ALtan24° = 26 tan 24° = 11.57mm
In triangle JGF: JF = JG tan 24° = 125 tan 24° = 556 mm
FC = JC  JF = 693  556 = 1.37mm
BF = LK  (LB + FC + CK) = 51  (1157 + 137 + 1625) = 2181 mm
GP = BF cos 24° = 2181 cos 24° = 2001 mm
PM = rad of small plug so that GM = 2001  8 = 1201 mm
GH = sum of plug radii = 125 + 8 = 205 mm
/\ GM
cos HGM = p^Y
Grl
1201 /\
1Zr W1 = 05858 and HGM = 54°8'
205
/\ /\ /\
Now HGN = PGN  HGM
and since PGN = 90° +24° = 114°
.HGN = 114  54° 8' = 59° 52'
GN = HG cos HGN = 205 cos 59° 52' = 205 x 05020 = 1029 mm
Distance h = [JG + GN + rad of small plug]  26
= 125 + 1029 + 826
= 479 mm
MISCELLANEOUS PROBLEMS IN MEASUREMENT 53
Example 23. For the turned part shown in Fig. 44, calculate the radius (/?)
so that the diameter at the throat may be 20mm as shown.
Fig. 44
The radius R blends into the 20° angular portion, and into the 20 mm
radius, spherical end.
We have to construct and solve an equation to give us R.
DC is drawn perpendicular to AB, and DF is drawn to the point where
R blends into the 20° angular portion so that FDE = 20° . GFE is parallel
to AB.
Then we have that GF + FE + CB = AB = 51 mm.
GF = GH cot 20° = (AH  AG) cot 20° = (AH 
= [AH  (CD  DE)] cot 20°
= [19  {(/? + 10)  R cos 20°}] cot 20°
= [19  R  10 + 09397/?] cot 20°
= [9 00603/?] cot 20°
= 2473 0166/?
FE = /? sin 20° = 0342/?
CE) cot 20 c
CB = VDB 2  DC 2 = V(R + 20) 2  (/? + 10) 2
Hence equating (GF + FE + CB) to AB = 51
2473  0166/? + 0342/? + V(R + 20) 2  (/? + 10) 2 = 51
and this reduces to
V(R + 20) 2  (/? + 10) 2 = 2627  0176/?
Square both sides:
(/? + 20) 2  (/? + 10) 2
(2627  0176/?) 2
54 MEASUREMENT AND GAUGING
i.e. by squaring out the brackets
R 2 + 40R + 400  R 2  20R  100 = 6901  9 25R + 003LR 2
Collecting and rearranging on the i?H side:
003 IR 2  2925.R + 3901 =*
This is a quadratic equation in R, and can be solved by the formula
method.
R =
2925 ±V2925 2  4 x 0031 x 3901
2 x 0031
From which R = 930 or 1355
The smaller root, R = 1355 is obviously the one we require
Hence R = 1355 mm
Exercises 2f
1. Calculate the diameter over wires for the following screw threads:
(a) M20 x 25 ISO metric over 0577/>mm wires
(b) M36 x 2 ISO metric over 0577pmm wires
(c) fin X lOt.p.i. Whitworth over 0564/? wires
2. A screw thread has the form and angle shown in Fig. 45. Calculate the reading W
over 8 mm plugs placed in opposite threads.
•*
J.
7&4
\Lj2
n pitch
Fig. 45
3. A wedge rests between two radiused jaws as shown in Fig. 46. When the top of the
wedge is level, calculate the distance H.
4. Solve the previous problem, when the radius on the righthand side jaw is 3 mm.
MISCELLANEOUS PROBELMS IN MEASUREMENT 55
5. In Example 3, if the wedge has an included angle of 40°, and is tilted through an
angle of 10°, calculate the distance H to its higher corner.
6. Calculate the distance from the centre of the circle (O) to the centre of the 20 mm
plug placed in the slot shown in Fig. 47.
Fig. 48
7. Fig. 48 shows the profile of a die form. From the information given, calculate the
width d at the narrowest portion.
Fig. 49
8. From the information given in Fig. 49 calculate the width W of the profile shown.
3 Calculations for cutting,
turning and boring
Speed and feed range
The reader will, no doubt, be acquainted with the meaning of speed and
feed in connection with turning and boring operations.
The driving arrangements of machine tools usually make provision
for a number of speeds and feeds, so that a suitable one may be chosen
for the work in hand. The reader will probably be curious as to how these
are determined. In the case of spindle speeds, the highest and lowest
speeds in the range are generally related to the extremes of size for which
the machine is designed. For example: a lathe might be designed to take a
rr.nge of work varying from 10 mm to 250 mm diameter. Allowing for a
cutting speed of 22 m/min, this would give for the top speed:
XT 1000 x 22 1000 x 22 x 7 _„ . .
N = T7 — = ^ 77j = 700 rev/min
n x 10 22 x 10
as being suitable for the 10mm diameter work.
For the lowest speed:
XT 1000 x 22 1000 x 22 x 7 „ . .
N = » X 250 = 22 x 250 = 28 rev/min
These will be the highest and lowest in the range, and if we assume there
are eight speeds altogether, six intermediate speeds must be chosen and
fitted in, the intermediate speeds being so calculated that the whole
series is in some regular order.
One method of arranging the speeds would be to make them in straight
line form, in which each speed would be the same amount greater than the
one below it. In this case, as there are 8 speeds and 7 intervals, each
interval would be:
Top speed — bottom speed _ 700 — 28 _ 672 _ _,
7 ~ 7  —  *&
SPEED AND FEED RANGE 57
and the speeds would be:
1st 28 rev/mm, 2nd 28 + 96 = 124 rev/min
3rd 124 + 96 = 220 rev/min
and so on.
700 
3 4 5 6
Speed Number
Fig. 50
If these were plotted on a graph, the result would be as shown in Fig. 50,
and the series is known as an Arithmetic Progression.
In practice, speeds arranged in this way are not suitable, as the steps
between the speeds at the lower end (28 rev/min, 124 rev/min, 220 rev/min)
are too great, whilst at the upper end of the range (700, 604, 508 rev/min,
etc.) a larger interval value could be tolerated without inconvenience.
To overcome these objections and provide a convenient range of
speeds, they are generally arranged in Geometric Progression. When
arranged in this way, instead of each speed being a constant amount
greater than the one below it, the speed is a constant multiple of the one
below it. The calculation for determining speeds arranged in geometric
progression is as follows:
Considering the case we have taken, where the extremes are 28 and
700 rev/min with 8 speeds.
The 2nd speed will be a constant amount multiplied by the 1 st, and the
3rd will be the same constant multiplied by the 2nd, and so on.
Let this constant be denoted by K.
Then
1st speed
2nd speed
3rd speed
4th speed
= 28
= 28 X K = 28AT
= 28*: x K = 2SK 2
= 2SK 2 x K = 2SK 3
and so on to the 8th speed, which we can see will be 2%K\
58 CALCULATIONS FOR CUTTING, TURNING AND BORING
Now the 8th speed is 700 rev/min, so that
K
2SK 1 = 700
28 ZD
No
 v/25 = 1584
25
Log.
7)1397 9(01997)
Hence we have 1 st speed = 28 rev/min
2nd speed = 28 x 1584 = 44 rev/min
3rd speed = 44 x 1584 = 70 rev/min, and so on.
The speeds arranged in geometric pregression are shown plotted in Fig.
5 1 , and their comparison with the straightline arrangement is shown in
the table below. The reader will observe that the geometric arrangement
gives closer intervals at the bottom of the range and wider ones at the
top which is more useful under application.
700
Arithmetic
Geometric
Speed
Progression
Progression
rev/min
rev/min
1
28
28
2
124
44
3
220
70
4
316
111
5
412
176
6
508
279
7
604
442
8
700
700
3 4 5 6
Speed Number
Fig. 51
We might put a geometric progression in general symbolic form as
follows:
Let
A x = 1st term
A 2 = 2nd term
A n = nth term
K = constant multiplier
n = number of terms
SPEED AND FEED RANGE 59
1 st term = A ,
2nd term = A 2 = A X K
3rd term = A 3 = A 2 K = A,K 2
nth term = A n = A X K"  l
(i.e. if n = 8, then the index of K is 8  1 = 7).
Hence if A n =A,K n ~ x
*» 1 = d* and £ = J4 5 
In an example, A, , A n and n would be given and the expression above
would enable K to be found and the whole series calculated.
Example 1 . Calculate a suitable range of six speeds for a drilling machine,
if the size range of the machine is to be from 25 mm to 10 mm drills and
a cutting speed of 22 m/min is to be given. Show a speed table with suitable
drill size for each speed.
IQO0S
We have that N = — 3— where jV = Spindle speed (rev/min).
S = Cutting speed (m/min).
d = Drill diameter.
For the top speed (25 mm drill)
„ 1000 X 22 1000 x 22 x 7 0Qnn . .
N * = *X2.5 = 22 x 25 = 2800rev/min
Lowest speed (10 mm drill)
„ 1000 x 22 1000 x 22 x 7 „_ . .
Nl = .xlO = 22 x 10 = 70 ° rev/mm
fa 72800
AT (the multiplier) =Jf =a/^ = >/4
No. Log.
4 5)0.6021(01204
Antilog 01204 = 1319 = K.
We may now calculate the range of speeds as follows:
1st speed = 700 rev/min
2nd speed = 700 x 1319 = 924 rev/min
3rd speed = 924 x 1319 = 1220 rev/min
60 CALCULATIONS FOR CUTTING, TURNING AND BORING
4th speed = 1220 x 1319 = 1610rev/min
5th speed = 1610 x 1319 = 2120rev/min
6th speed = 2120 x 1319 = 2800rev/min
The relationship between the speeds and the drill sizes they will accom
modate may now be calculated:
1st speed, 700rev/min suitable for 10 mm drill
700
2nd speed, 924rev/min suitable for 10 x q~j
700
3rd speed, 1220rev/min suitable for 10 x
4th speed, 1610rev/min suitable for 10 X
5th speed, 2120rev/min suitable for 10 x
1220
700
1610
700
2120
Nearest 025 mm
10mm
75 mm
575 mm
425 mm
6th speed, 2800rev/min suitable for 25 mm drill
The speed and drill diameter table is shown below.
325 mm
2 5 mm
Spindle Speed (rev/min)
700
924
1220
1610
2120
2800
Suitable drill diameter (mm)
100
75
575
425
325
25
Feeds
The factors governing the choice of feedrange limits is rather beyond
our scope, but when the limits of the range have been fixed, together
with the number of intermediates in the range, the steps usually follow
the rules for geometric progression in the same way as the speeds.
Exercises 3a
1. A lathe is operating on a range of work varying from 25 mm to 250mm diameter.
Allowing for a cutting speed of 22m/min, calculate the highest and lowest speeds
necessary. If there are 8 speeds in the complete range, find the range of speeds if they
are in geometric progression. Make out a table showing the most suitable diameter to
be turned on each speed.
2. A drill has 4 speeds and drills a range of holes from 2 mm to 6 mm diameter.
Calculate the four speeds if they are in geometric progression, and make a drill diameter
speed table. Cutting speed = 165m/min.
CUTTING TOOL LIFE 61
3. For Ques. 1 p. 60 plot a graph showing spindle speeds vertical, and speed number
horizontal.
4. On a 400mm stroke singlepulley, allgeared shaping machine the number of
turns of the driving pulley required to make one complete double stroke of the ram
were found to be as follows:
1st speed, 27 turns
2nd speed, 16 turns
3rd speed, 8 turns
4th speed, 4 turns
a • *u *• Cutting time 125
Assuming the ratio =— — ^—. to be — ; — and to remain constant, estimate a
Return time 1 '
suitable pulley speed to give an average cutting speed of llm/min in the lowest gear
and on the longest stroke.
With this pulley speed, find the most suitable stroke for each of the other speeds.
5. The highest spindle speed for a small lathe is 1500 rev/min. In order to obtain a
suitable cutting speed for drilling some 3 mm holes in brass, a drill head is mounted
on the carriage, and driven in the opposite direction to the spindle. At what speed
must the drilling spindle be driven to give a cutting speed of 66 m/min?
Cutting tool life
As a cutting tool does its work it becomes blunted, and a time ultimately
arrives when it must be taken out and resharpened. The life of the tool
between the times of regrinding is influenced by the severity of its
treatment whilst it is cutting. Depending upon circumstances, there is a
best economic tool life for every tool; if the cutting duty is made such as
to allow the tool to last longer than the best economic time, then it is
cutting below an efficient rate and is doing less work then it might. On
the other hand, if its performance is raised to a level such as to cause
it to become blunted in less than the economic time, then undue ex
pense and lost time are being incurred in the additional sharpening and
resetting necessary.
The problem of tool life and of cutting generally is rather complicated
and indeterminate, since there are so many variable factors involved.
From experimental data, however, cutting speed and tool life have been
found to conform roughly to the following rule:
VT» = C,
where V = Cutting speed in metres per minute
T ~ Corresponding life in minutes
C = A constant depending on cutting conditions
62 CALCULATIONS FOR CUTTING, TURNING AND BORING
\ to \ for roughing cuts in steel \
for roughing cuts in cast iron > i
for light cuts in steel )
— 12
_ J_
— 10
for roughing cuts in steel j ~ :j ^ ;„„,„
carbide tools.
The above values of n are only approximate and are influenced by
tool shape, use of cutting compound, etc. The above relationship enables
us to estimate probable tool life as shown in the following examples:
Example 2. When operating with roughing cuts on mild steel at 20m/min
a certain tool gave a life of 3 hours between regrinds. Estimate the life of
this tool on similar cuts at a speed of 30m/min.
In the VT" = C expression for this case we will take n = £.
The first step is to calculate the value of the constant C.
We have that when V = 20, T = 180 min and n = \.
Hence C = 20 x 1801
Taking logs No. log
T „ , _ log 180 180 8)22553
LogC=log20+ 8~ 180r ^02819
= 15829 20 13010
15829
Antilog 15829 = 3827 = C.
Hence we may write:
We now require T when
VT* = 3827
V= 30
307^  3827
T k = 2*£Z = 1.2757
T = (12757) 8
T = antilog 08456
= 7008, say 7 min.
No. Log
12757 01057
8_
08456
TOOL CUTTING ANGLES 63
Example 3. A tool cutting at 20m/min gave a life of 1 hour between re
grinds when operating on roughing cuts with mild steel. What will be its
probable life when engaged on light finishing cuts? [Taken = £ for rough
ing, and fo for finishing cuts.]
Here we have for roughing:
No.
log
20 x 60i = C
60
8)l7782(
C = antilog 15233 = 3338
60i
02223
20
13010
15233
Applying to the finishing conditions
20 x n =
3338
n  3 ^ 8  1669
T = (1669) 10
No.
log
1669
02225
(1669) 10
2225
T = antilog 2225
 1679, say
168 min.
Tool cutting angles
The principal angles on a cutting tool are its rakes and clearances, and
these are shown in Fig. 52.
Side Rake
Front (Top)
Rake
Front
Clearance
Side
Clearance
Fig. 52
The choice of suitable cutting angles depends upon the material being
cut, and the reader should look up particulars in books dealing with
64
CALCULATIONS FOR CUTTING, TURNING AND BORING
Workshop Technology. We might consider here the effect on the cutting
angles of various tool settings.
When a tool is cutting a circular bar of material it is operating relative to
a radial line drawn from the centre of the work to the cutting point. Thus
in Fig. 53(a) the tool is cutting relative to the line OA. If the tool point is
is level with the centre of the work the line OA is horizontal and the cutting
angles operating are the true values put on the tool.
If, however, the tool point is placed above or below the centre, the
cutting angles will be modified since the line OA is not now horizontal.
Top Rake (r)
i_
Clearance(c)
Fig. 53(a).
Top Rake
^■Clearance
Fig. 53(6).
In Fig. 53 (b) the tool point is shown a distance h above the work centre.
The line OA is now tilted up an angle a, and if R is the radius of the work
we have that sin a = =
K
The top rake angle of cutting will be increased by the angle a, and since
the total angle /5 has not changed the clearance will be reduced by a. If the
tool is put very high, the clearance will be so reduced that it will vanish
altogether and the tool will rub instead of cut. If the tool is placed below
the centre the effects are opposite, the rake being decreased and the
clearance increased. We might calculate the effect on the cutting angles of
a numerical example.
Example 4. A bar of material 60 mm diameter is being turned with a tool
having 20° top rake and 6° front clearance. Calculate (a) the cutting
angles when the tool is 25 mm above centre, and (b) the amount the
tool must be above centre for the clearance to become zero.
TOOL CUTTING ANGLES 65
The conditions are as shown at Fig. 53 (b).
(a) sin ^^^J^ 00833.
From which a = 4° 47'.
The rake is increased and the clearance decreased by this amount.
Thus the rake becomes 20° + 4° 47' = 24° 47' and the clearance be
comes 6°  4° 47' = 1° 13'.
(b) If the clearance is to vanish, then angle a must be 6°.
sin a = ^7jand since sin 6° = 01045
01045 = *j
h = 30 x 01045
= 314mm
Exercise 3b
1. A certain tool when cutting cast iron had a life between regrinds of 2 hours when
cutting at 20m/min. If the relationship between life and speed is given by VT™ = C, cal
culate C, and estimate the tool life at a speed of 15m/min.
2. For the tool in Question 1, plot a graph of tool lifecutting speed over a range of
speeds from 30m/min to 15m/min.
3. For a certain tool it was found that the relationship between speed and tool life was
given by VFi = 50. Estimate the cutting speed to give a time of 2 hours between re
grinding.
4. If the relationship for highspeed steel tools is VT* = C„ and for tungsten carbide
tools VT 5 = Q, and assuming that at a speed of 25m/min the tool life was 3 hours in
each case, compare their cutting lives at 35m/min.
5. For a certain tool it was found that the relationship between cutting speed (K)and
tool life (T) was as follows:
Express T in terms of V and find T when V = 25m/min.
6. A cutting tool has a top rake of 20° and a clearance of 7° . Calculate the modified
values of these angles when the tool is cutting 3 mm below centre on a bar 44 mm diameter.
7. When turning a bar 50 mm diameter, how much above centre may a tool with a
clearance of 6° be set before the clearance vanishes? When the tool is in this position, what
is the effective value of the top rake, if the rake on the tool is 15°?
8. A boring tool 10mm deep is required to bore out holes to 40mm diameter. If the
body of the tool is horizontal, how much clearance will be necessary if the bottom corner
of the tool is to clear?
66 CALCULATIONS FOR CUTTING, TURNING AND BORING
9. If the clearance face of the tool in Question 8 is made with two angles as shown
in Fig. 54, calculate the angles a and /5 for the tool to clear the hole.
Fig. 54
10. A 50 mm diameter boring bar is concentric with a 72 mm hole which is being bored.
The tool is 12mm square and passes through the centre of the bar. Calculate the tool angles
necessary so that the cutting rake shall be 10°, and the clearance 6°.
Taper turning
When a tapered or formed surface is being turned the accuracy of the sur
face is influenced by the position of the tool point relative to the centre
of the work. The conditions in the case of taper work are shown in
Fig. 55 (a) . If the tool is set on the centre, and its movement controlled so
as to turn the correct taper, it starts at A, and when it has travelled the
line AB, the correct taper has been produced.
End View
Fig. 55
If, now, instead of being on the centre, the tool is a distance h below
centre and starts at C, it will travel along the line CD, and if D is on the
same circle as B, the length CD is greater than AB. The tool being set to
TAPER TURNING 67
move the distance AB will therefore not reach the point D, and con
sequently if A and C are on the same circle, the top diameter of the taper
will be less than it should be. The reader will probably appreciate the
point better if he considers the extreme case of the tool being at E and
travelling parallel to AB. In such a case, if the tool could cut in this posi
tion, it would not turn a true taper, but would produce a tapered shape
faintly hollow in form.
In order to follow the problem mathematically we will show an enlarged
diagram (Fig. 55 (6)).
The tool starts at C and moves out the distance CF, equal to AB. If it
turns the end of the bar to radius r, instead of reaching the radius OB = R
it will only attain OF = R l .
Let the work be L long as shown in Fig. 55 (a).
True taper = * — p — *
Actual taper obtained = *■ ' ~ — 
An attempt to reduce R , to an expression in terms of the other quantities
involves rather awkward terms, and cases will be best evaluated from the
information available . Such an example is illustrated as follows:
Example 5. In turning a taper of 1 in 6 on the diameter, the tool is set to
move the correct angle relative to the work, but is 4mm below centre. If
the work is 32 mm diameter at the small end, calculate the actual taper
obtained.
Referring to Fig. 55 (b), if we assume the work to be 60 mm long, then
the large diameter will be 32 + 10 = 42 and we shall have
r = 16 mm
R = 21mm
h = 4mm
The tool, starting at C where OC = 16 mm, will move out to F where
CF = 21mm.
We require the distance OF.
OF 2 = FG 2 + OG 2 = (FC + CG) 2 + OG 2
But FC = 5 and OG = 4
.. OF 2 = (5 + CG) 2 + 4 2
68 CALCULATIONS FOR CUTTING, TURNING AND BORING
But CG 2 = CO 2  OG 2
= r 2  h 2 = 16 2  4 2
240
CG = v / 240~ = 1549
Hence OF 2 = (5 + 1549) 2 + 4 2
= 2049 2 + 4 2 = 4358
OF = \/43T8~= 2088
The top diameter of the taper will therefore be 2 x 2088 = 4176 instead
of 42mm.
Since the bottom diameter is 32 mm and length 60 mm
Actual taper = 4176  32 = 9^6
H 60 60
This gives a taper of 1 in 615.
Form tools
For turning forms from the crossslides of turret and automatic lathes and
sometimes from centre lathes as well, a tool is used which gives the
correct form on the work. If the tool is set on centre as shown in Fig. 56,
it must have the correct form in plane OAB. Since, however, the tool
must have clearance as shown, lengths such as BC, taken perpendicular
to the front clearance face will be less than lengths such as AB taken
on the horizontal, and the form of the tool on a plane parallel to BC will
be different from its form on AB . When the tool is being made, the shaping
and other machining operations are carried out parallel to the clearance
face, so that for the purpose of making the tool it may be necessary to
determine its form when taken on a plane such as BC, perpendicular to
the front clearance face.
U
^Clearance Fig. 56
FORM TOOLS 69
We shall best illustrate the method of determining the modified form of
the tool in planes perpendicular to the front clearance face by working one
or two examples:
Example 6. A tool is to have the form shown in Fig. 57 (a) on its top
horizontal face. If the clearance is 10°, determine and sketch the form on
a plane perpendicular to the front clearance face.
15 10 72
^ i.
55 
(a)
11 2
15
10 7;
i
^
p
/
V
r
55 »
e)
(
IF2>
Fig. 57
In Fig. 57 (a) horizontal dimensions will not be affected, but vertical
BC
dimensions will be shortened in the ratio of jtt in Fig. 56, and since
/\ gp
ABC is equal to the clearance angle, — — = cos of clearance angle
AB
= cos 10° = 09848
Hence the 112 in dimensions become 112 x 09848 = 1103 mm and
the 45° angle will become a triangle as shown in Fig. 57 (b).
tan A = ^ = 10154. From which A = 45° 26'
1103
Hence the revised sketch of the tool profile taken along the clearance
face will be as shown in Fig. 57 (c).
When the form of the tool incorporates circular shapes the problem
becomes rather more involved, since the effect of shortening the depth
but not the width converts the circular form into a portion of an
ellipse.
70 CALCULATIONS FOR CUTTING, TURNING AND BORING
Example 1 . Calculate and sketch the form of the tool shown in Fig. 58 (a)
when taken perpendicular to the front clearance of 12j° .
As before, the vertical dimensions are shortened in the ratio of the
clearance angle cosine, i.e. cos 12° = 09763.
We thus have 128 x 09763 = 1250 mm
8 x 09763 = 781 mm
and for the 15 mm radius,
15 X 09763 = 14.64mm
The base of the 20° angle is shortened to 09763 of its original length,
so that if we divide the tangent of 20° by 09763 we shall have the tangent
of the modified angle
tan 20° = 0364
Q976 3 = 03728 = tan of modified angle = tan 20° 27'
The sketch of the modified profile is shown at Fig. 58 (b) .
125 116 30 88 6
1
28 8
^C20 c
iM
75
(a)
125 116 30 _8_8 6
Elliptical Form
Fig. 58
The finishing of the elliptical form given to the circular portion is apt to
be troublesome, but such a shape can be produced on a grinding wheel
by trimming it with a radius forming attachment set off centre. This is
shown in Fig. 58 (c), and if the radius truing attachment is set with the
FORM TOOLS WITH "TOP RAKE" 71
diamond rotating in plane CDE it will trim the wheel to a semicircle,
15 mm rad., in that plane. Since AB is less than CD, the true form of the
wheel on a radial plane such as ABO will be elliptical, because the width
of the wheel at B = width at D. The semiminor axis of the elliptical
form will be AB = FD, and the major axis will be 2 x 15 mm = 30 mm.
As the profile on a radial plane is the one imparted by the wheel to the
work we can, by forming the wheel in this way, obtain the required
elliptical shape for the tool in question and we require to determine h
in order that when CD = 15 mm, FD will be 1464 mm.
Let us assume a grinding whell of 200mm radius, and consider the
problem from the aspect of two intersecting chords of a circle.
Then FD.DH = CD.DG
But FD = 1464, DH = 400  1464 = 38536 and CD = 15
Hence (1464) (38536) = 15.DG
„ 14 64 x 38536 „, , mm
DG = rr = 3761 mm
and CG = 3761 + 15 = 3911
3911
CE = ±CG = ^ = 19555
But h 2 = R 2  CE 2 = 200 2  19555 2 = 1760
from which h = 42 mm
Hence by trimming the wheel to a 15 mm radius in a plane 42 mm off
centre, the required elliptical form will be produced.
Form tools with "top rake"
If back slope is put on the cutting face of a form tool the cutting effect of
the tool is rather curious, because for the purpose of obtaining an accurate
relative reproduction of the tool form on the work, the tool, at the finish
of the cut, must have its top face lying on a radial line. This is shown at
Fig. 59 (a), where the tool is shown at the completion of its cut, and its
top face lies on the radial line OA. The reader will notice that at this
position the effective top rake is zero, since the tool is cutting relative to
tangent BC, which is perpendicular to OA.
When this tool starts its cut, however, the conditions are as shown at
(b), and it will be seen that if the back slope a on the tool is made large
72 CALCULATIONS FOR CUTTING, TURNING AND BORING
enough, the tool will start cutting with an effective top rake of/3 = a  S.
As the tool feeds in, this rake will gradually get less until, as we have seen
above, there is zero rake at the final position. The reader will notice that
a tool of this type must be set below the centre by the amount h, where
h
 = sin a and r = smallest radius being turned.
The calculation for the modified form to which the front face of such
a tool should be made is similar to that we have already dealt with, except
that the angle a must be taken into account.
In Fig. 59 (c), a is the backslope angle and c the clearance.
The tool finishes its cut relative to face AB, but the depths of the form
are put on parallel to AD (i.e. along CB).
Hence a length AB on the top of the tool will correspond to CB, per
pendicular to the clearance face.
In triangle ABC: C = 90° and A^Sb = a + c
CB = AB cos (a + c).
Hence depths in the form must be shortened in the ratio of cos (a + c).
CIRCULAR FORM TOOLS
73
Fig. 60 Circular Form Tool.
Circular form tools
On some types of automatic lathes forming is done from the crossslide by
means of a circular form tool, a sketch of which is shown at Fig. 60.
These tools have the advantage that the form may be turned on their
rim and they may be used all round the rim by continual resharpenings.
Cutting clearance is obtained by making the cutting edge AB some
distance h below the centre, and the tool is applied to the work as shown
at Fig. 61 . The clearance angle a is then the angle CAO and
sin a =
AO Tool radius (r)
and h = r sin a
Fig. 61
74 CALCULATIONS FOR CUTTING, TURNING AND BORING
Gashing the tool in this way results in a variation between the form
turned on it and the form it imparts to the work, because a radial depth
DB on the tool will turn a depth AB on the work. Widths on the form are
unaffected, and corrections for depths may be calculated as follows,
where d = a depth on the tool and / = corresponding depth on work.
From the property of intersecting chords of a circle
But
and
Hence
DB.BF = ABBE.
DB = d; BF = 2r  d; AB  /
BE = AE — / = 2r cos a — I
d(2r  d) = l{2r cos a  I)
which reduces to a quadratic equation in d as follows:
d 2  2rd + l(2r cos a  I) =
in which all the quantities except d are known.
The following example will illustrate the application of this.
Example 8. A circular form tool, 100 mm diameter, is to be made to
produce the form shown in Fig. 62 (a). If the gashing is to give a cutting
clearance of 10°, determine the form to be turned on its periphery.
75r^
m
425
(b)
Fig. 62
Here r = 50 mm and a = 10°, so that
h = 50 sin 10° = 50 x 01736 = 868 mm.
To turn the correct diameters on the work, the steps on the face AB
of the tool must be:
325  175 __ ,325  125 ,.
■= = 75 mm, and ■= = 10 mm
Applying the equation above to these two cases, we have
CIRCULAR FORM TOOLS
75
(0 / = 75, r = 50, cos a = cos 10° = 09848
d 2  lOOtf  7.5(100.09848  75) 
from which d = 738 mm
(«) / = 10, r = 50, and cos a = 09848
d 1  lOOrf  10(100.09848  10) =
from which d = 982 mm
The angle on the tool to give an included angle of 1 20° on the work must
now be corrected . Its width is
325  125
tan 30°  10 tan 30° = 5774 mm
and its corrected depth from (ii) above is 982.
Hence
5774
noy = tan of its angle — 05882 from which the angle is found
to be 30° 28'.
The turned profile of the tool is shown at Fig. 62 (b).
Exercises 3c
1. A lathe centre is being ground up by a 70 mm diameter grinding wheel fixed to a
toolpost grinder on the compound slide. If the centre of the wheel is set 10mm below
the axis of the lathe centre, and the compound slide fed at 30°, find the angle to which
the centre will be ground.
2. A lathe is set correctly for turning a taper of! in 6 when the tool is on the centre.
What taper will be produced on apiece ofwork 120 mm long and 60 mm top diameter when
the tool is 5 mm below (jentre?
3. A form tool is straight, and set at an angle of 15° with the axis of the work (i.e. to
form an included angle of 30° on the work) . If the tool is set to the above angle, but 5 mm be
low centre, calculate the actual angle produced on a job 50 mm top diameter and 40 mm long .
?7Z7
&.
I
Fig. 63
4. A form tool having 8° clearance and 15° back slope is required to turn the diameters
shown at Fig. 63. Calculate (a) the depth AB on the clearance face of the tool, (b) the
amount the tool should be below centre, (c) the top cutting rake at the commencement of
cutting. (Top diameter ofwork = 10mm).
76
CALCULATIONS FOR CUTTING, TURNING AND BORING
5. Calculate the tailstock setover for turning a taper of 8° included, on a job 115 mm
long. What variation in taper will be produced by a variation of ±25 mm in the length
of the work?
6. Calculate the depths and the angle on a form tool for producing the form shown in
Fig. 64. Tool has no top rake and 10° clearance.
Fig. 64
7. At Fig. 65 is shown a thread form which is to be finished by the form tool indicated.
Calculate the angle, depth and bottom land as measured from the clearance face of a
tool having no top rake and 15° clearance.
524
Fig. 65
8. A circular form tool 55mm diameter is gashed to give 10° clearance, and is to turn
the form indicated at Fig. 63. Calculate the amount offcentre for the gashing and the
depth AB on the tool.
9. Calculate the depths and angle to be turned on a circular form tool 60mm diameter
to give the profile shown at Fig. 66, if the tool is gashed to give 12° clearance.
APPROXIMATE CHANGE WHEELS FOR ODD THREADS 77
375^ \876
Fig. 66
10. In cutting the thread on a core for a diecasting mould, the pitch of the thread
is to be made 2% longer than standard to allow for contraction. To do this, the tailstock
of the lathe is set over, the lathe taper attachment is set parallel to the work, and then
the thread is cut in the usual way. Calculate:
(a) The actual pitch required if the nominal thread is 2mm pitch.
(b) The tailstock setover if the core is 150 mm long between centres.
(c) The angle to which the taper attachment must be set.
Calculating approximate change wheels for odd threads
Sometimes a case may arise where an odd thread must be cut, the exact
pitch of which cannot be obtained with the standard machine change
wheels. Also, if a lathe is not supplied with the special 127T wheel, and a
metric pitch is required, some alternative way of getting a suitable pitch
becomes necessary.
The method of continued fractions will often provide a very near ratio
to that required, and enable a pitch to be cut which is near enough for the
purpose.
In Appendix VII it will be seen, that the convergents of a continued
fraction are a series of fractions, each succeeding one approaching closer
to the true value of the original ratio. If, therefore, we have a complicated
ratio, the exact value of which cannot be obtained on the machine, it
is quite possible that by converting to a continued fraction and finding
the convergents, one of these convergents will be a ratio that can be
used, and its value will probably be close enough for the purpose. The
method will be best illustrated by examples.
78 CALCULATIONS FOR CUTTING, TURNING AND BORING
Example 9. Find the nearest pitch to 2 18 mm that may be cut on a lathe
with a 5 mm leadscrew, and give suitable change wheels from a set ranging
from 20T to 120T in steps of 5T.
^, . r Drivers .„ , 218 mm 109 , . irkn .
The ratio of gears: p^ will be — E = ^~r, and since 109 is a
6 Driven 5 mm 250
prime number, the exact ratio could not be obtained without a gear of
this size.
Converting this to a continued fraction and finding the convergents we
have:
109)250(2 The continued fraction is:
218 I
32)109(3 2+1
96 3 + 1
13)32(2 2+1
26_ 2 + 1
6)13(2
11
1)6(6
and the convergents are: 1st = \, 2nd = f, 3rd = &, 4th = & 5th = $§.
If we take the 4th convergent (#) we may obtain a gear ratio as
follows:
n _ 2 x 85 _ 20 x 85 Drivers
37 ~ 6 x 65 ~ 60 x 65 Driven
To find the actual pitch obtained we must multiply the ratio $ by the
pitch of the leadscrew, i.e. $ X 5 = ff, which when converted to a
decimal gives 2 1795 mm. This is less than 003% in error on the required
pitch of 2 18 mm.
Example 10. Find the nearest pitch obtainable to 2\ mm, on a lathe with
a 6 t.p.i. leadscrew and a set of wheels as in the last example.
2Jmm
The pitch of the leadscrew is £in and the ratio required
Converting the inches to millimeters (lin = 254 mm)
2\ 6 9x6 27 270
= 24 x
nn
'254 * 254 4x254 508 508
POWER REQUIRED FOR CUTTING 79
The continued fraction is
270)508(1
270 1 + 1
238)270(1 1 + 1
238 7+1
32)238(7 2+1
224 3 +
14)32(2
28
4)14(3
11
2)4(2
and the convergents: 1st = }; 2nd = ±; 3rd = &; 4th = #; 5th = ft;
6th — 2io
Ulll — 508
The 4th convergent is the last one which can be made into a ratio and
gives:
17 = 2 x 85 20 x 85 Drivers
32 ~ 4x8 ~ 40 x 80 Driven
The actual pitch obtained will be £ x £ X 254 = 22489 mm, being
0001 1 mm short.
Power required for cutting
Turning and Boring. When metal is being cut with a singlepoint tool as
in turning and boring, the tool is subjected to pressure in three directions
at right angles: (1) vertical chip pressure, (2) horizontal work pressure
across the lathe, (3) horizontal feeding pressure along the lathe.
The first of these is of greatest importance from the aspect of the
power absorbed. The other two, although absorbing some power, are of
small effect when compared with the vertical pressure and are generally
neglected.
From numerous experiments that have been made it has been esta
blished that the cutting force on a single point tool is connected in an
expression of the form
F = Cd a f b
where F = force; d = depth of cut;/ = feed, and C = a constant.
a and b depend on the metal being cut and other factors.
80 CALCULATIONS FOR CUTTING, TURNING AND BORING
For most practical purposes the expression
F = Kdf= K(Cut area) gives results good enough.
K is a constant depending on the metal being cut.
If S ~ cutting speed in metres per minute, the work done per minute
F x S
will be F x S, and the power — ^ — Watts.
Hence Power = 60 qq Q Kilowatts
Approximate values for K are as follows:
Metal Steel Steel Steel Steel r „ t A ,
being 100150 150200 200300 300400 f^T Brass Bronze 2\„
cut Brinell Brinell Brinell Brinell ™*~
K
(N/mm 2 ) 1200 1600 2400 3000 900 1250 1750 700
[From the form of the expression, the reader will observe that K is the
force on the tool per square millimetre of cut area.]
When the power required to do the cutting has been calculated, the
total power to run the cut and overcome friction in the machine may be
found by adding on about 30%.
Example 1 1 . Calculate the power being absorbed in running a cut 3 mm
d^ep with a feed of 15 mm, on a mild steel bar 50 mm diameter turning
at 140rev/min.
. n X 50 x 140
Cutting speed = 
1000
22 x 50 x 140
~ 7 x 1000
If we take the constant K as 1200
= 22m/min
D 1200 x 3 x 15 X 22
Power = 60000 = 198kW
Adding 30% for frictional losses in the machine we have
198 + 198 Xi= 198 + 0594 = 2574 say, 25 kW to run the machine .
POWER FOR DRILLING
81
Power for drilling
When a drill is cutting it has to overcome th e resistance offered by the
metal and a twisting effort is necessary to turn it. This effort is called
the Turning Moment or Torque on the drill . The units for torque are those
of a force multiplied by a length and the most usual is the Nm unit. The
turning effect of a force, or a pair of forces, acting at a certain radius,
is found by multiplying the force by the radius, or for two forces the
turning effect is the sum of the product of each force by its radius. Thus
in Fig. 67, if the drill required a torque of lONm to turn it, the torque
would be equivalent to equal and opposite forces of 250 N each operating
at 20 mm radius. [T = 250 x 002 + 250 X 002 = lONml
SON Resistance at
each corner
Fig. 67
In addition to the torque, a drill requires an axial force to feed it
through the work, but in power calculations this is generally neglected.
When the torque is known, the work done is found by multiplying it by
the number of turns made and by In .
Thus if T = torque in Nm and N = speed in rev/min,
the work done per minute = In NT Nm
, , 2nNT , „ r
and the power = ^^ kw
82 CALCULATIONS FOR CUTTING, TURNING AND BORING
The torque required to operate a drill depends upon various factors,
but for the purpose of being able to obtain an approximate calculation for
it we will omit all but the drill diameter, the feed and the material being
drilled. The relation between the torque, the diameter and the feed has
been found experimentally to be that torque varies as/ " 75 /) 1 ' 8 .
Using this, we may say that
Torque (T) = Cf° 15 D hS newton metres
where C = a constant depending on the material
/ = drill feed (mm/ rev)
D = diameter of drill (mm).
When the torque has been found, the power can be calculated as shown
above.
The following table gives approximate values for the constant (C).
A , . . . , . , .,, , Alu Soft Cast Steel Carbon
Material being drilled , , ., , s ,
minium brass iron (mild) tool steel
C 011 0084 007 036 04
Example 12. Calculate the power required to drill a 20 mm hole in mild
steel at 250rev/min and a feed or 05 mm/ rev. Find also the volume of
metal removed per unit of energy.
Taking the constant, C, from the table as 036 we have
7 , = 0.36/° 75 Z) 1  8 Nm
/ =05
D = 20
T = 036 (05)° 75 (20) 18
and taking logarithms.
log T = jog 036 + 075_log 05 + 18 log 20
= j5563 + 0.75(16990) + 1.8(13010)
= 15563 + 075(03010) + 18(13010)
= 04437  02258 + 23418 = 16723
T = antilog 16723 = 47 Nm
Since the speed is 250rev/min
_ 27r.250.47 _
60 000 ~ 1ZJKW
POWER FOR TURNING AND DRILLING 83
Volume of metal removed per minute
= (Area of hole) (Feed) (Speed)
= j x 20 2 x 05 x 250 = 39 275 mm 3
Energy consumption = — mow = 319mm 3 /watt minute.
= 053mm 3 /joule
Exercises 3d
1. Calculate the nearest change wheels for cutting a sparkingplug thread (lmm pitch)
on a lathe with a 4 t.p.i. leadscrew and a set of wheels ranging from 20T to 120T in steps
of 5T. For the ratio you select, find the actual pitch of thread obtained.
2. A worm having a lead of Timm is to be cut on a lathe with a 5 mm leadscrew. Taking
n as 31416, express the ratio required as a continued fraction, and find the nearest
convergent that can be used with a set of wheels specified for the last example. What was
the actual error in the lead obtained for the worm?
3. A shaft revolves at 15 rev/min and requires a thread cutting on it which will cause
a nut to move along the shaft at 665mm/min, when it turns at the above speed. Find the
lead of the thread required and calculate the nearest that can be cut to it on a 5 mm lead
screw with change wheels specified for Question 1. What is the actual speed of the nut
with the thread you obtain?
4. Estimate (a) the power input to a lathe when it is taking a 6 mm cut in cast iron
with 075 mm in feed at 20m/min. (b) The volume of metal removed per unit of
energy. Take the overall efficiency of the machine as 70%.
5. A lathe is just able to run a cut of 5 mm depth at a feed of 08 mm in steel of
120 Brinell, at 150 rev/min on a 50 mm diameter bar. Estimate what cut could be taken
on 25 mm bars of 250 Brinell material, at 240 rev/min and the same feed as before.
[Take values of K from the table on p. 80.]
6. Taking the value of K from the table on page 80, estimate what cut could betaken
on a lathe turning bronze b ars at 06 mm feed and 20 m/min, if 5 kW were available, and 30%
of the power were lost in friction.
7. For the lathe in Question 4, estimate the power cost per 8hour day, with power at
2p per kWh and the efficiency of the motor is 80%.
8. Calculate the torque required to drill 20 mm diameter holes in mild steel, at a feed
of 025 mm/ rev. If the drill speed is 300 rev/min, calculate (a) the power absorbed in cutting,
(b) the energy absorbed per cubic millimetre removed per minute.
9. A 25 mm drill is drilling aluminium at HOm/min. Calculate its speed. If the feed is
03 mm/rev, calculate the torque, and the input power if frictional losses are equivalent to
30% of the cutting power.
10. For the drill in Question 8, calculate the drilling time for 100 holes, each 40 mm deep.
If the electrical efficiency is 80%, calculate the cost to drill these 100 holes using the same
electrical data as in Question 7.
11. By using a continued fraction calculate the nearest set of change wheels to cut the
thread specified in Example 10(a) Exercises 3 c, leadscrew 5 mm pitch.
4 Calculations for gears and
gear cutting
Formatiion of the involute tooth
For various practical and theoretical reasons, the tooth shape most
commonly used for gearing is the involute. Before we commence our con
sideration of various problems connected with involute teeth, it will be
as well to examine the involute curve itself.
Involute
If a cord is wrapped tightly round a circular form and then unwound, at
the same time being kept tight, the end of the cord will trace out an
involute. This is shown at the top of Fig. 68, where CB is the portion of
the cord that has been unwound and AC is the involute. Another method
Involute
Involute
Straight
edge
Fig. 68
of tracing an involute is to roll a line (e.g. a straightedge) on a circle
when the end of the line will trace out an involute. The size of circle will,
of course, influence the shape of the involute, but there are certain
properties which are common to all involutes. Those properties which
are interesting from the aspect qf the involute as a tooth form, are as
follows:
INVOLUTE TOOTH FORM
85
(1) A tangent to the involute is always perpendicular to a tangent from
the same point on the involute to the circle from which it is formed. In
Fig. 68 the cord CB is tangential to the circle at B, and a tangent to the
involute at C is perpendicular to CB.
(2) The length of the cord CB is equal to the length of the arc AB.
Tooth form
O, and 2 are the centres of a pinion and gear of which the pitch circles are
shown tangential at P, which is called the pitch point .
AB is a line through P perpendicular to line C^Cv CD also passes
through P and is included at the angle if> to AB. CD is called the line
of action and $ the pressure angle. The name "line of action" is given
to CD because it is on, and along that line, that the pressure between
the teeth takes place. The angle (j> nowadays is virtually always 20°, but
at one time 14£° was more common, being related to half the inclined
angle of the Acme thread form . With the changeover of the pressure angle
from 14£° to 20° the clearance between the top of the tooth of one gear
and the base of its mating tooth form has been increased from 5% of the
circular pitch ( = 0157 of the module), to 025 of the module. The cutting
depth for gears of 20° pressure angle is thus 225 times the module. To
Fig. 69 Formation of Involute Tooth.
86 CALCULATIONS FOR GEARS AND GEAR CUTTING
obtain the tooth shape a circle is drawn tangential to CD called the base
circle. The portion of the tooth between this circle and the top (EF) takes
the form of an involute to this circle; the portion of the tooth below this
circle (FG) is radial (i.e. on a line joining the end of the involute to the
gear centre). This is all shown in Fig. 69, where for the sake of clearness
the construction has only been carried out on the lower gear. The con
struction also only shows one side of a tooth; the other side is merely the
same shape reversed, and spaced away, a distance equal to the tooth
thickness.
The involute rack
The rack is a gear of infinite diameter so that its pitch circle will be a
straight line (AB, Fig. 70). A base circle of infinite diameter tangential
^ F Fig. 70
to the line of action CD will be a straight line coinciding with CD. The
involute to this will be the straight line EP, and the radial continuation
will be PF. Hence the side of the rack tooth is straight, and inclined at
the pressure angle. The complete tooth will have a total angle equal to
twice the pressure angle (Fig. 70).
In the following considerations of gear elements the following symbols
will be used for the quantities stated:
No.ofteethingear ... Tor r Pressure angle (ji
Diametral pitch P Module m
Circular pitch p Addendum of tooth . . Add.
Diameter of pitch circle Dord Dendendum Ded.
Radius of pitch circle . R or r
The recommended manner of quoting the size of a gear tooth form,
irrespective of the units employed, is to quote the addendum and refer
to it as a module. In which case,
pitch circle diameter = module x number of teeth
or D = mT
THE TOOTH VERNIER
87
The module is therefore the reciprocal of the diametral pitch, irrespective
of whether measurements are made in millimetres or inches. All the
formulae which follow can be used for diametral pitch, by substituting
— for m. Until the use of the metric module becomes the preferred usage
in describing gear tooth sizes, the reader may find gears listed in a diamet
ral pitch series. The diametral pitch is simply the number of teeth per
unit of pitch circle diameter, and consequently it is necessary to be par
ticularly careful in stating that unit, e.g.
diametral pitch of 8
or diametral pitch of 02
(inch series)
(mm series).
In any case, conversion to a module is simply effected from the relation
ship that the module is equal to the addendum; furthermore
circular pitch p = n x module m
The tooth vernier
The geartooth vernier is an instrument for measuring the pitch line
thickness of a tooth. It has two scales and must be set for the width (w)
of the tooth, and the depth (h) from the top of the tooth, at which w occurs
(Fig. 71).
Fig. 71
The angle subtended by a half tooth at the centre of the gear (AOB) in
Fig. 71).
. ,360 90
= iof^=^=
88 CALCULATION FOR GEARS AND GEAR CUTTING
m> w *ri ■ 90 z> • 90
AB = y = AO sin ~y = R sin =■
D = module x number of teeth
D = 2R = mT
a n mT
and R = j
„ w _ . 90 mT . 90
Hence ^ = i? sin = = z sm =
A T ■ 90
and w = wisin=r
To find fc we have that h = CB = OC  OB.
But OC = R + Add. =^ + m
A ™ Z> 90 W7, 90
and OB = R cos = = = cos ~.
TT r mT mT 90 .... ,, — ■ ^
Hence /* = y + /w = cos ^ = m + ^ 1  cos ^ \. (2)
mrf 90l
T[ !  CQ8 TJ
Example 1 . Calculate the gear tooth vernier settings to measure a gear
of 33 T, 25 metric module
^ . 90 ._ ,, . 90
w = mT smjr = 25 X 33 sin^
= 825 sin 2°43' = 825 X 00474
= 391 mm
, mT.. 90,
h = m + 2~(1  cos yr)
= 25 + 2 ' 5 * 33 (1  cos2°43')
= 25 + 41.25(00011)
= 2545 mm
Constant chord method
One drawback to the method just outlined is that the measurements w
and h depend on the number of teeth (T) in the gear, and for each dif
ferent gear a fresh calculation has to be made. The following method
avoids this and gives a constant pair of readings for all gears of the same
pitch and pressure angle.
CONSTANT CHORD METHOD 89
Fig. 72
In Fig. 72 is shown a gear tooth meshing symmetrically with a rack.
O is the pitch point, and as we have seen above, the gear tooth will contact
with the straightsided rack tooth at the points B and D, lying on the
line of action.
Then DB = w will thus be constant for all teeth of the same pitch and
pressure angle.
Since EOA is the pitch line of the rack
nm
and
EA = j circular pitch = \p = y
OA = !EA = ™
A A
In triangle OAB: B = 90° and O = <f>
:. OB = OA cos i/j,
and in triangle OCB: C = 90° and B = ^
.\CB= OB cos ij).
nm
Hence CB = OA cos iji cos ijt = OA cos 2 i/j = —r cos 2 iff
and
7tm
(3)
DB = 2CB = w = j cos 2 (/)
h = Add.  OC = m  OC.
OG = OB sin ip and OB  OA cos iff.
OC = OA cos (/t sin $ = j cos ^ sin ^
nm . . . f~ 7t , • ,1
— j— cos (j> sin iff = m\ 1 — j cos $ sin ty j
It will be seen that expressions (3) and (4) remain constant if module
(m) and pressure angle (^) do not alter.
But
Hence
and h = /
(4)
90
CALCULATION FOR GEARS AND GEAR CUTTING
Example 2. Calculate the constant chord and the depth at which it occurs,
for a 30T gear of 6mm module, 20° pressure angle.
Here we have
h> = —x cos 2 20
sin 20° = 0342
cos 20° = 09397
6 X 3142
X (09397) 2 = 832 mm
m(l — j cos ij) sin f)
= 6(1  07854 x 09397 x 0342) = 6 x 07476 = 449 mm
Plug method of checking for pitch diameter and divide of teeth
The tooth vernier gives us a check on the size of the individual tooth,
but does not give a measure of either the pitch diameter or the accuracy
of the division of the teeth.
Fig. 73
Fig. 73 shows a rack tooth symmetrically in mesh with a gear tooth
space, the curved sides of the gear teeth touching the straight rack tooth
at the points A and B on the lines of action. O is the pitch point. If now
we consider the rack tooth as an empty space bounded by its outline, a
circle with centre at O and radius OB would fit in the rack tooth and touch
it at A and B (since OA and OB are perpendicular to the side of the rack
tooth). Since the rack touches the gear at these points, the above circle
(shown dotted) will rest against the gear teeth at points A and B and will
have its centre on the pitch circle.
In triangle OBD: OB = radius of plug required.
CHECKING FOR PITCH DIAMETER AND DIVIDE OF TEETH
91
OD = \ circular pitch =
A A
B = 90° O = ijf.
Tim
7im
OB = OD cos if/ = j cos (jf
nm
Dia of plug = 20B = y cos iji
(5)
This is the diameter of a plug which will rest in the tooth space and have
its centre on the pitch circle. Notice that the plug size remains the same
for all gears having the same pitch and pressure angle.
With such plugs placed in diametrically opposite tooth spaces, it is a
simple matter to verify the gear pitch diameter. The accuracy of the
spacing over any number of teeth may be found as shown in chordal
calculations.
Example 3. Calculate for a 36T gear of 5 mm module and 20° pressure
angle, (a) plug size (b) distance over two plugs placed in opposite spaces,
(c) distance over two plugs spaced 10 teeth apart.
Tim ^S7T
(a) Dia of plug = ^ cos ^ = ~ cos 20° = 7854 x 09397
= 738 mm
Pitch dia of gear = mT = 5 x 36  180mm
(b) Distance across plugs in opposite spaces
= 180 + 738 = 18738 mm
(c) Distance across plugs spaced 10 teeth apart (Fig. 74).
Fig. 74
92
CALCULATION FOR GEARS AND GEAR CUTTING
'lf l {\
Angle subtended by 10 teeth = 10 x ~y = 100° . In triangle OAB:
AB = OA sin 50° = 90 x 0766 = 6894
Centre distance of plugs = 2 x AB = 2 x 6894 = 137 88 mm
Distance over plugs = 13788 + 738 = 14526 mm
Base pitch
The base pitch is the circular pitch of the teeth measured on the base
circle. It is useful for checking the angle between adjacent teeth and for
checking a tooth against "drunkenness."
Fig. 75
If AB (Fig. 75) represents a portion of the base circle of a gear, and
CD and EF the sides of two teeth, and then the length FD is the base
pitch. But if any lines such as CE and HG are drawn tangential to the
base circle cutting the involutes at the points shown, then
EC = GH = FD
If there are T teeth in the gear, then T x FD = 2ttR b and
FD
iTlRt
If ij) is the pressure angle, then from Fig. 69, page 85,
H0 2 = rad. of base circle (R B ) = P0 2 cos ij> = R cos tjt.
BASE PITCH 93
Hence FD = base pitch = ,£° S ^
n , 2nR TtD
But ~y~ = ~f = 7tm
( D \
(since = = m )
Hence base pitch = nm cos <ji (6)
This is the distance between the curved portions of any two adjacent
teeth and can be measured either with a height gauge or on an enlarged
projected image of the teeth.
Exercises 4a
1. Determine the diameter of a plug which will rest in the tooth space of a 4mm module
20° rack, and touch the teeth at the pitch line. Calculate (a) the distance over two such
plugs spaced 5 teeth apart, (b) The depth from the top of the plug to the top of the
teeth.
2. Calculate the gear tooth caliper settings for measuring the following gears:
(a) 37T, 6mm module; (b) 40T 20mm circular pitch.
3. A 5 mm module involute rack tooth is measured at its pitch line and found to be
799 mm wide. If the tooth spacing and angle are correct, what error has been made
in the cutting of the teeth? (Pressure angle = 20°.)
4. A 30T replacement gear of 5 mm module is required, and the nearest cutter
available for cutting the teeth is one of 5 diametral pitch, (inch series). If the blank
is turned to the correct module dimensions, and the cutter sunk in to the depth marked
on it, what will be the error in the tooth?
5. Determine the "constant chord" dimensions for the following gears:
(a) 8 mm module, 20° pressure angle; (b) 25 mm circular pitch, 20° pressure angle.
6. Calculate the diameter of plug which will lie in the tooth space of a 5 mm module
gear with its centre on the pitch circle. If the gear has 50T, find (a) distance over two such
plugs spaced in opposite spaces, (b) distance over two plugs spaced 12 spaces apart
(u = 20°).
7. Two plugs are placed in adjoining spaces of a 29T gear 20mm circular pitch, and
the gear is stood up resting on them. Calculate the distance from the face upon which
these plugs are resting, to the top of a similar plug placed in the tooth space at the top
of the gear. [Press, angle = 20°, and plug diameter is that which rests with its centre
on the pitch circle.]
8. Determine the base pitch of the following gears: (a) 30T 8 mm module, ij> = 20°;
(b) 30T, 25 mm circular pitch, ij> *= 20°.
9. Two teeth of a 30T gear of 125 mm module, 20° pressure angle are projected to
a magnification of 50. Calculate, to the nearest 05 mm the length on the projected image
of the following measurements:
(a) base pitch,
(b) depth of tooth space,
(c) chordal thickness of tooth at pitch line,
(d) height from root of tooth to pitch line.
94 CALCULATION FOR GEARS AND GEAR CUTTING
Stub teeth
For some purposes, particularly when gears are subject to shock and
vibration, the tooth of standard proportions is apt to be weak and liable
to break. In such cases stub teeth are often used.
For this type of tooth the size is indicated by a fraction. The numerator
of the fraction expresses the module to which the circumferential pro
portions of the tooth conform, and the denominator determines the
radial proportions of the tooth. Thus a £ stub tooth means one in which
the pitch diameter is worked out on a basis of 5 mm module, and the
tooth height on the proportions of 4 mm module. Since 5 mm module gives
a larger normal tooth than 4 mm module, the result is a short stubby tooth,
hence the name. Stub teeth are generally cut with a pressure angle of
20°, and the following are the pitches most commonly used:
5 6 8 10 125 , 15
and
4' 5' 6' 8 ' 10 125
The same method is used when quoting stub tooth sizes in diametral
pitches, but in this case the ratio produces a "proper vulgar fraction",
i.e. with the numerator smaller than the denominator. Hence, if a stub
tooth is denoted by a fraction in which the numerator is smaller than the
denominator, the reader should appreciate that the sizes quoted refer to
diametral pitches and not modules, and it will be necessary to state
whether the diametral pitch is "millimetre series" or "inch series".
Example 4. Calculate the principal dimensions for a 45T gear having a
f stub tooth.
Here: the pitch diameter will be worked out on 6 mm module and the
heights on 5mm module.
Pitch dia = mT = 6 X 45 = 270mm
Addendum = m = 5 mm
Top dia of gear = 270 + 2(5)
= 280 mm
Cutting depth = 225 X 5
= ll25mm
Backlash in gearing
If a pair of gears were cut theoretically correct and assembled at the
correct centre distance, a tooth on one gear would just fit hard into the
tooth space of the other, because the pitch line width of the tooth and
MEASUREMENT OF BACKLASH
95
space would be equal. For freedom of action the above conditions would
be unsuitable, and it is usual to allow a little play between the thickness
of the tooth and the width of the space into which it fits. This play is
called "backlash," and it is the backlash which allows one gear to be
turned a fraction before the drive is taken up by the mating gear.
The amount of backlash to be allowed depends on the tooth size, and
the following table gives an indication of suitable allowances:
Module (mm)
10
8
6
5
4
3
25
2
15
Backlash [in mm
clearance between
face of mating teeth]
04
04
04
03
02
015
015
01
01
Measurement of backlash
Two suitable methods of measuring backlash are (a) by means of feeler
gauges between the teeth, (b) by measuring the distance that the centres
of the gears may be moved nearer together from the standard distance
before the teeth are in hard contact.
The first of the above methods is straightforward and needs no mathe
matical manipulation. For the second method it will be helpful for us to
obtain an expression giving the backlash in terms of the amount the gears
are capable of being moved together.
Fig. 76
In Fig. 76, ab is half the backlash and P is the pitch point.
If we consider the portion cb of the tooth as being a straight line, then
triangle abc is rightangled at b, ac is the amount the gear centers may be
moved together, ab = \ backlash and angle acb = t/i.
96 CALCULATION FOR GEARS AND GEAR CUTTING
Let B = backlash = lab
and D = amount the gear centres can be moved together = ac
Then — = sin ib
ac
B_
2
D = sin ij) and B = 2D sin (b
B
or D =
2 sin (Ji
Helical (spiral) gears
The type of gear we have dealt with so far has been the spur gear, i.e. one
in which the teeth are straight and parallel with the axis of the gear. In
helical gears the teeth are not straight, but are cut on a helix. These gears
are also called spiral gears, screw gears and skew gears.
A sketch of a portion of a helical gear is shown in Fig. 77, in which it will
be noticed that the teeth slope at an angle a (the helix (spiral) angle) to the
axis of the gear, and in the gear shown, the teeth are cut RH helix. Teeth
may, of course, be RH or LH.
Referring again to the figure, it will be seen that the pitch of the teeth
may either be taken round the rim of the gear, or it may be taken perpendi
cular to the teeth. In the first case it is called the circumferential pitch
ip c ), and in the second case it is known as normal pitch (p„). Circular pitch
is still considered as being taken along the surface formed by the pitch
circle. The circular pitches p n and p c are shown in Fig. 77, and as will be
seen, the relation between them is the same as the relation between the
A A
sides AC and AB of triangle ABC. In this triangle C = 90° and A = helix
angle a .
Hence £*■ = — — = cos a,
Pc AB
i.e. p n = Pc cos a (7)
Pn
or p c = r = p n sec a
rc cosct rn
It will be seen that if a section is taken through the teeth on a plane con
taining AC, we shall have the true shape of the tooth as it is cut. Hence the
normal pitch (p n ), which is the one measured parallel to AC, is the pitch
which governs the cutter to be used to cut the gear . Referring again to Fig .
HELICAL (SPIRAL) GEARS 97
[Measured orn
[Pitch Circle]
\
Fig. 77.
77, the pitch p c , multiplied by the number of teeth, gives the pitch circum
ference. This pitch, then, governs the size of the gear, which does not
depend on the number of teeth. The larger the helix angle <r, the greater
will be the ratio — and the larger the gear for a given number of teeth.
Pn
As well as circular pitch p„ and p c we may, just as in the case of spur
gears, express the pitch in the module form . The normal module (m n ) is the
one which governs the true shape of the tooth, and since cutters are most
commonly specified in terms of the module, this is important to us from
the point of view of cutting the gear. The relation between /w n and/>„ is the
same as for spur gears.
i.e.
Pn = *™n
(8)
For a spur gear we have D = mT
For a helical gear, since the circular pitch
Pc =
Pn
COS<7
andD =
, then m c =
mT
cos a
cos a
m
cos a
(9)
98
CALCULATION FOR GEARS AND GEAR CUTTING
Addendum of the tooth = m
and the cutting depth (working depth + clearance)
== 2 addendum + clearance = 2 m + clearance.
(10)
In modern practice, using a 20° pressure angle, the cutting depth has been
standardised at (working depth + 025 addendum)
==2°25/m (11)
Helix (spiral) lead and angle
The teeth of a helical gear are cut on a helix, which is merely a screw thread
with a very large lead. The reader will, no doubt, be aware that a screw
thread can be developed into a triangle, and the only differences between
the development of a geartooth helix and that of a screw thread are: (a)
the helix angle of a thread is the complement of that for a wheel tooth (Fig.
78(a) [complement of an angle = 90°  the angle] ; (b) the development is
' Helix ^^"
Fig. 78 Development of Helix.
based on the assumption that the helix makes one complete turn round the
cylinder upon which it is cut. In a screw thread this is true, but the tooth
of a gear only completes a small proportion of a complete turn. [The
student may imagine a helical gear to be a short length of a very coarse
CUTTER FOR HELICAL GEARS 99
thread having as many "starts" as there are teeth in the gear.] On the
assumption noted under (b) above, the development of a helical gear
tooth is shown in Fig. 78(6). From the diagram we have the following
relationships
Pitch circum nD
tan a = = j = p
Lead L
, , _ nD nmT sec a .,,.
or lead L = = — (12)
tan a tan a
Cutter for helical gears
When cutting spur gears with a form cutter on the milling machine, the
cutter is marked with the range of teeth for which it is suitable. For
example, to cut a 30T gear, we should use a No . 4 cutter, which is suitable
for a range of 26T to 34T. Due to the twist on the teeth of a helical gear this
must be modified, and the size of cutter is given by
No. of teeth (T) ( .
(cos a) 3 ( '
Example 5. The pitch diameter of a helical wheel is to be approximately
120 mm, the helix angle is 30° and it is to be cut with a cutter of 4 mm
module. Find the particulars of the nearest gear to this.
We have that the normal module = 4 mm and a = 30°.
4T
Hence, from (9) 120 =
and T =
cos 30°
120 cos 30°
4
= 30 x 0866 = 2598
The nearest to this is 267
, n mT 26x4 104 1oni
and D = , ^r = = 1201 mm
cost? cos 30° 0866
Top dia. = 1201 + 2 Add = 1201 + 2(4)
= 1281 mm
Cutting depth = 225m = 900 mm
v mom a fur 7lD 3142x1201 ,..
From (12) Lead of helix = ^r = ttt^ = 654 mm.
tan 30° 0577
100 CALCULATION FOR GEARS AND GEAR CUTTING
Thus the nearest gear has the following particulars: 26T. Pitch dia.
1201 mm, Helix angle 30°, Lead of helix, 654mm.
96
The cutter for this gear would be ,_ , „ , = 40, i.e. the same cutter as
(UoOOJ
would be used for a 40T 4mm module, spur gear.
Example 6. Two parallel shafts at 120 mm centres are to be connected by
a pair of helical gears to give a speed ratio of 1:2. The helix angle of the
wheels is to be approximately 20° . If the normal module is 25 mm, deter
mine suitable wheels.
Since the shafts are parallel, the helix angles of the two wheels will
be the sarnie, but one will be RH helix, and the other LH
Centre distance =120
Ratio of speeds = ratio of pitch radii =1:2
r = i x 120 = 40 and R = f X 120 = 80
Hence D = 160 and d = 80, also m = 25 m
f , v n mT „ D cos a 160 X 09397 „ . 
from (9) D = , T = = == = 6013
cos a m 25
Since the ratio is to be 2: 1 let us try 60T and 30T, and find a new
value for a.
D cos a
Then for the wheel from T =
m
mT
cos a ■ =
25 X 60 150
~ 160 ~ 160
= 09375
hence a = 20°22'
To obtain the leads of the helices of the wheels we have for the
wheel:
Pitch cir cum = 160tt = 5027 mm
5027 5027
and from (19) Lead = tan 2Qo22 , = ^^ = 1354mm
For the pinion Lead =  —  Oo 'y = half of lead of wheel = 677 mm
HELICAL GEARS 101
The particulars of the gears are thus as follows:
Wheel No. of teeth = 60 Pinion. No. of teeth = 30
Pitch dia = 160 mm Pitch dia = 80 mm
Add = 2 5 mm Top dia = 85 mm
Top dia = 165 mm Cutting depth = 5625 mm
Cutting depth = 225 m Helix angle = 20°22'.
= 5625 mm LH helix
Helix angle = 20°22'. Helix lead = 677 mm
RH helix. ~ ., + 30
Helix lead = 1 354m Cutter t0 USe= T053tSJ>
_ 60 = cutter as for 36T spur
Cuttertouse = (093W wheel.
= cutter as for 73T spur
wheel.
The reader will observe that the solution to these problems is arrived
at by a compromise after a system of trial and error. With wheels of this
type such a procedure is nearly always necessary before a practical set
of working conditions can be arrived at. Generally, the conditions allow
one or more of the gear elements to be varied to suit the problem.
Exercises 4b
1. Calculate the pitch diameter, top diameter and cutting depth for a 42T gear having
J stub teeth.
2. Find the geartooth caliper settings for checking the tooth of a 32T, J stubtooth
gear.
3. A pair of gears are required to connect two shafts at 160 mm centres. Ifthe speed ratio
required is 3:5, and the gears are to have £ stub teeth, find their leading particulars.
4. A pair of gears consists of 27T and 63T gears of 4mm module, 20° pressure angle.
If the backlash allowance is 02 mm, what should be the centre distance between the two
gears when the teeth are hard in contact?
5. A train of gears consists of the following: 30T driving 48T driving 48T driving 7 5T.
The teeth are 4 mm module and the backlash allowance on all the teeth is 01 5 mm. If all
the backlash is taken up in one direction, through what angle must the 30T gear be turned
before the drive is taken up by the 75T wheel?
6. Calculate the following particulars for a 52T spiral gear of 4 mm normal module,
20° spiral angle: (a) pitch diameter, (b) top diameter, (c) cutting depth, (d) lead of spiral,
(e) suitable cutter to use.
7. A helical gear is to have a helix angle of30°(RH), and the normal module is 4 mm. The
pitch diameter must be as near as possible to 125 mm. Calculate (a) the number of teeth,
(b) the pitch, and top diameters, (c) the lead of the helix, (d) the correct cutter to use.
102
CALCULATION FOR GEARS AND GEAR CUTTING
8. A spiral gear has 25 teeth, a helix angle of 45°, and an approximate pitch diameter of
75mm. Calculate the nearest normal metric module. Find also the pitch diameter, top
diameter., lead of spiral and cutting depth.
9. A pair of spiral gears are required to connect two parallel shafts 100 mm apart, with
a gear rat io of 3 : 2. Working on a normal module of 25mm, and an approximate helix angle
of 20°, determine particulars of a suitable pair of wheels.
Worm gearing
A worn drive is often used to connect two noninteresting shafts which
are at rightangles and a fair distance apart. The worm is the equivalent
Lead(L)
Start
Af?7
Starts
N°. s 2&3
Start
N92
Lead 'Angle (/{.)
Fig. 79 Diagram of 3Start Worm.
of a screw thread, the shape of a section of the thread on a plane through
the axis of the worm, being the same as a rack tooth in the involute system .
The action between a worm and wormwheel is equivalent to that of the
wheel as a gear; rolling along the worm as a rack. As the worm is usually
produced by turning, or by a milling process similar to turning, the pitch
Fig. 80 Worm Thread Form for 14£° Pressure Angle.
WORM WHEEL 103
most commonly used is the circular pitch (p) . Fig . 79 shows the pitch, lead
and lead angle for a worm, and Fig. 80 gives the proportions for the thread
on a section through the axis for a tooth of 14° pressure angle.
The relationships between the pitch, lead and lead angle are the same as
for a screw thread.
Lead (L) = (Axial pitch)(No. of starts) = pn
Lead L
tan A =
n (Pitch dia) nd
The pitch diameter of a worm is quite an arbitrary dimension, and a
worm may be cut to any pitch diameter suitable to accommodate it to
centre distance at which it is to engage with the wheel. In general, the pitch
diameter should not be less than four times the pitch.
Worms with large lead angles
The efficiency of a worm drive increases with the lead angle up to a maxi
mum at about 45° . In view of this, multistart worms with large lead angles
are to be preferred. Unfortunately, with such worms, interference dif
ficulties occur in cutting and operation, and when the lead angle exceeds
20° it is usual to increase the pressure angle (i.e. the included angle of
the worm thread is increased). The pressure angle may be taken up to 20°
(40° incl. angle of thread), and in very quick start worms iji is sometimes
made 30° . For pressure angles other than 14£° the tooth proportions must
be recalculated on the basis of the new angle and will not be the same as
those shown on Fig. 80. Also, when the lead angle exceeds 15°; it is more
advantageous to base the tooth proportions on the normal pitch. For the
case of the tooth shown in Fig. 80 these are modified as follows:
Normal pitch = (axial pitch)(cos of lead angle)
p n = p cos X (as (7) above)
Then Addendum of thread = 0318/»„
Depth of thread = 06866/?,,
Width of finishing tool at bottom = 03 \p n
Worm wheel
For the best results the crosssection of the wormwheel rim should be of
the form shown in Fig. 81 (a). For light duty and moderate speeds, wheels
as shown at 81 (b) will give satisfactory results. Teeth with curved bot
104
CALCULATION FOR GEARS AND GEAR CUTTING
Wor m Cen tre
Wheel Axis
(CL)
mm
Fig. 81
toms should be cut with a hob or a fly cutter; that at 81 (c) is really a helical
gear and may be cut as such. The shape of the teeth of worm wheels is the
same as for involute gears of the same pressure angle.
Worm wheel dimensions
These will be based on the circular pitch and for large worm lead angles
the tooth proportions will be in terms of the normal pitch. In reading the
following, Fig. 81 (a) should be referred to:
Pitch dia (D) =
Pitch circum.
IE
71
Throat dia = Pitch dia + 2 add  D + 063/)
Centre distance (C) = Pitch rad wheel + pitch rad worm
_D d
~ 2 + 2
Throat rad (R t ) = C  \ throat dia
Whole dia = 2(C OA)
= 2[C*,cos(i0)]
Width (w) approx = 2BC
= 2 (Top rad of worm) [sin (£/J)]
= (d + 0636/?) sin (i/J)
In cases where tooth proportions are based on normal pitch, then in
expressions (19) and (23) above, p n should be used instead of/?.
SPEED RATIO WITH WORM GEARING 105
Speed ratio with worm gearing
If the wheel has T teeth, and the worm is single threaded (1 start), then
the worm will turn T times for 1 turn of the wheel. A 2start worm will turn
T
■yr times per revolution of the wheel, and so on. Hence for T teeth
in the wheel and n starts on the worm:
Rev of worm T
Speed ratio =
1 Rev of wheel n
Example 7 . Determine the dimensions of a worm and wheel to operate
at 120 mm centres and give a ratio of 16 — 1. Circular pitch 10 mm, pres
sure angle 14£°, wheel face angle 75°.
For 10 mm pitch the minimum dia. of the worm should be 4 x 10 mm
= 40 mm. This leaves 120 — 20 = 100 mm as the radius of the wheel. Dia
of wheel = 200 mm. Circum = 200tt = 6284 mm.
At 10mm circular pitch this gives 63 teeth.
If we make the worm 64T and use a 4start worm, we shall obtain the
16—1 ratio required.
Pitch dia wormwheel = = = 2037 mm
71 71
2037
and pitch rad = — ~ — = 10185 mm
Throat dia = 2037 + 0636/)
= 2037 + 0636(10) = 21006 mm
Throat rad = 120  K2 1006) = 1497 mm
Whole dia = 2tl20  1497 cos 37±°]
= 2 [120  1497 x 07949 = 21624 mm
Pitch rad worm = centre distance — rad wheel
= 120  10185 = 1815
and pitch dia = 2 x 1815 = 3630 mm
Top dia worm = 3630 + 0636/? = 3630 + 636 = 4266 mm
We may now find the approx. width of the wheel (h>)
w = (d + 0636» sin 37±° = 4266 sin 37£° = 4266 x 06088
== 2597 mm (say 26 mm)
tan of worm lead angle = — r = ^t^k = 03508
° Ttd 7i x 3630
X = 19° 20'
Whole depth of tooth = 06866/> = 6.87 mm
Width of threading tool at end = 031/? = 3 10 mm
106 CALCULATION FOR GEARS AND GEAR CUTTING
This gives us all the data necessary for the worm and wheel.
As an exercise, the reader should make a fullsize drawing of the pair.
Helical (spiral) gears to mesh with worms
Sometimes, as a compromise, and to avoid the delay and expense of
ordering a new wormwheel, an ordinary straightfaced helical wheel
is cut to replace a wormwheel. The main problem is to find, out of stock,
a suitable cutter to give a satisfactory matchup.
If stocks of circular, diametral and module pitch cutters are available,
it is often possible to arrive at a workable solution. Let us see what could
have been done to match a helical wheel to the worm in Example 7 above:
Normal circular pitch of wormwheel = 10 cos 19°20'
= 10(09436) = 9436 mm
9436
Normal module = = 3003
71
This is very near a 3 mm module and a 3 mm module cutter would
probably cut a satisfactory gear.
Bevel gears
Bevel gears are used to connect two shafts whose axes meet, and which
are in the same plane. We saw in connection with spur gears that the
motion of two gears was equivalent to that of two thin cylinders or discs
rolling together, the diameters of the discs being the same as the pitch
diameters of the gears. In the case of bevel gearing the fundamental
conception of the motion is that of two cones rolling together.
Fig. 82
BEVEL GEAR CALCULATIONS 107
In Fig. 82, OA and OB represent the axes of two shafts intersecting
at O. COD and DOE are the two elemental cones, having OA and OB
as their axes. The cones touch along the line OD, and if one is turned it
will drive the other. In practice, the gears only consist of a narrow
frustrum of the cones and are shown thickened. Metal is added at the
back, as shown, to strengthen up the teeth in that region.
The two elemental cones are called the pitch cones and become the
imaginary pitch surfaces of the gears. The angle 6 P is the pitch angle of the
pinion, W that of the wheel and I is the shaft angle. When teeth are
cut in the gears, the number of teeth in each gear will be proportional to
the pitch diameters CD and DE.
Hence if n = rev/min of wheel and N = rev/min of pinion
n T DE sin W
N t CD sine,
Note that 6 P + W =Z
Bevel gear calculations
/\
In Fig. 83: AOB = 6 is the pitch angle.
AC = pitch dia (/)).
/\
DOA = a is the addendum angle
S\
AOF = /5 is the dedendum angle
/\
EOB = $ = 6 + a is the face angle
/\
FOB = p = 6 — ji is the root angle
C is the cone distance.
L is the tip distance.
C
/is the face width and may be made about = ~
D, L and C are given capital or small letters according to whether
they refer to the wheel or to the pinion. The angles are generally given
a suffix p or w to differentiate them, [p for pinion and w for wheel.]
The back face DAF is always made perpendicular to the pitch surface
AO, and the size and shape of the teeth as developed round that face
correspond to the proportions for the pitch of the teeth in the gear
108 CALCULATION FOR GEARS AND GEAR CUTTING
Add m
Fig. 83
(e.g. if the gear had teeth of 5 mm module, then the teeth at the surface
DF (as shown developed round line CH) would be the correct size for that
pitch). In travelling down the tooth from face DF towards O, every line
on the tooth converges to O .
From Fig. 83:
=  C = sin 6 and C = ^—. — —
2 2 sin 6
tan a
tan =
Add m , . j i x
— pr = ^ (where m = module)
Ded _ l25m
C ~ C
Whole diameter (over corners DE) = pitch dia + 2AD cos 6
But AD = Add = m
Whole dia = D + 2m cos
Examples 8. Two shafts at 90° are to be geared together by bevels. The
speed ratio is to be 3:2 and the pinion pitch diameter is to be 120mm.
Determine the dimensions of 5 mm bevels.
Since the ratio is 3:2 and d = 120mm
D = 120 x I = 180mm T = 36 and t = 24
BEVEL GEAR CALCULATIONS 109
Fig. 84
Signifying the pitch angles by 6 W and p .
Then tan P = ^ (Fig . 8 3). =% = \
6 p = 33° 41' and B w
Add = m = 5 mm Ded
90  33° 41' = 56° 19'
125/n = 625 mm
OA = V90 2 + 60 2 = 108 10 mm
5
tan of add angle
tan ded angle
1081
625
1081
= 00463 a = 2° 39'
= 00578 /$ = 3° 18'
Face angle wheel <t> w = 56° 19' + 2°39' = 58° 58'
Face angle pinion <j> P = 33°41' + 2° 39' = 36° 20'
Root angle wheel P w = 56° 19'  3° 18' = 53° 1'
Root angle pinion P p = 33° 41'  3° 18' = 30° 23'
Whole dia (wheel) = D + 2m cos W = 180 + 10(05155) = 185 16 mm
Whole dia (pinion) = d + 2m cos P = 120 + 10(08056) = 12806 mm
Face width (if made }C)
/ =
120
20
6 x 05155 05155
A sketch of these wheels is shown at Fig. 84
'2 sin
= 3880 mm
110 CALCULATION FOR GEARS AND GEAR CUTTING
Shafts not inclined at 90°
When the shafts are inclined at angles other than 90° the calculation for
the pitch angles is slightly more difficult. The following example will
indicate a method of evaluating such cases.
Example 9. Determine the dimensions of 4 module bevels to connect two
shafts at 70°. The pitch diameter of the pinion is to be 80 mm and the
ratio 4: 5.
From the information given we have
d = 80 mm
D = 80 X  = 100mm
4 4
Z, = 70°
In triangles OAC and OAB (Fig. 85), AC = 50 and AB = 40
Also
Hence
7TT = sin 6 W and ~r = sin 6 P = sin (70 c
OA OA
50 . .
OA = Sin ^
M
^=sin(70<
K)
3*39'
(1)
(2)
Fig. 85
BEVEL GEAR CALCULATIONS 1 1 1
40
sin (70 c
>
K)
50
sin
9 W
=
sin 70°
cos
o w 
 cos 70° sin W
sin
e w
09397
cos
o w
0342 sin W
sin
o w
sin W
Divide (2) by (1) ^ =
sui u w
sin 70° ens ft — ms 70° sin ft
(see Appendix VI)
0937
0342
tan W
4 0937
^ + 0342 = f^4r from which e w = 30° 22'
5 tan W w
and p = 70°  Ow = 70°  39° 22' = 30° 38'
nA AB 4° 1QAQ
OA = sin 30° 38' = 0^095 = 7848mm
Add = module = 4 mm
Ded = 4 x 125 = 5 mm
4
tan add angle = ^^ = 005107 a = 2° 55'
/o4o
tan ded angle = ^^ = 00638 /J = 3° 39'
Face angle wheel W = 39° 22' + 2° 55' = 42° 17'
Face angle pinion p = 30° 38' + 2° 55' = 33° 33'.
Whole dia wheel = D + 2m cos W = 100+8 cos 39° 33' = 106 18 mm
Whole dia pinion = D + 2m cos P = 80 + 8 cos 30° 38' = 8688 mm
The main dimensions are shown on Fig. 85.
Exercises 4c
1. Calculate the top diameter, root diameter and helix angle for a 2start worm of
10mm pitch and 40mm pitch diameter.
2. If the worm in Question 1 drives a wheel, and the gear ratio is 17£ to 1, calculate
the centre distance, and full particulars for the worm wheel.
3. A worm is 60mm pitch diameter, and 20mm circular pitch. Find the number of
starts in order that the lead angle may be approximately 30°, and state the actual
lead angle. Calculate the normal thickness of the thread on the pitch line.
4. A worm drive operates at 150 mm centres and the ratio required is 15 to 1. Taking
a circular pitch of 10mm and a worm about 50mm pitch diameter, find a suitable worm
and wheel for the drive.
5. Two shafts at 90° are to be connected by equal bevel wheels. Determine the
dimensions of one of these if the module is 25 mm, and the pitch diameter 60mm.
1 12 CA LCULATION FOR GEARS AND GEAR CUTTING
6. Two shafts at 90° are to be connected by bevels to give a ratio of 3:2. If the pitch
diameter of the smaller bevel is 130 mm and the module is 5 mm, determine the
dimensions of the wheels.
7. A bevel gear has 20 teeth of 4mm module and a pitch angle of 45°. Determine
the top diameter and included angle of a taper pin, which will rest in a tooth space
with its centre on the pitch cone, its large end level with the back face of the gear, and
its curved surface making contact with the tooth sides for its whole length. (ij> = 20°).
8. Two shafts, inclined at 120°, are to be connected by bevel wheels to give a ratio
of 4:3. The pitch diameter of the smaller gear is 90mm and the teeth are 5mm module.
Determine particulars of the gears.
5 Milling and the
milling machine
Milling Cutters
For the purpose of considering the calculations necessary in connection
with milling cutters, we may divide them into three general types: (a) those
with fluted teeth, (b) machine relieved, (c) inserted teeth. Sketches of
these are shown in Fig. 86.
(a) fluted Cutter
(b)Machine Relieved
Cutter
Fig. 86
(c) Inserted Blade
Cutter
Number of teeth
Milling cutters and milling conditions vary so widely that it is difficult
to set hard and fast rules for determining the number of teeth to be put
in a cutter.
For fluted and relieved cutters the rule N = 275 VD  58 gives a
reasonably proportioned tooth [N = No. of teeth, D = Diameter of
cutter] .
The formula N = pr + 8 gives a fairly coarse tooth for cutters over 60 mm
diameter.
Take a 100 mm cutter, the first formula gives:
AT = 2.75\/lOO  58 = 22 teeth
whilst the second expression gives
N = }W + 8 = 16teeth
114
MILLING AND THE MILLING MACHINE
For an insertedbladeface mill it is better to assess the number of teeth
on the assumption of their being spaced a suitable distance apart on the
periphery of the cutter. Thus if we take a 200mm face mill and assume
the blades to be spaced 30 mm apart, we have
No of blades =
Circumference of cutter 200;*
30
30
= approx20
Rake on cutter teeth
If a cutter has its teeth milled with their faces radial, and parallel to the
cutter axis as shown in Fig. 87(a), the teeth have neither top nor side rake .
If, however, the end view of the tooth is as shown in Fig. 87(6) the front
rake is the angle a, since the tooth cuts relative to a radial line OA from
the centre to its tip. Side rake is put on the tooth by milling it on a helix
as shown at Fig. 87(c). The side rake is then equal to the helix angle /3.
When teeth are of this form the relation between the hand of the helix and
the direction of rotation of the cutter is important, in order that the end
thrust introduced may be accommodated efficiently.
(CL)
No Top Rake
No Side Rake
(c) Side Rake
Fig. 87.
The calculation of the helix angle is discussed in the section on spiral
milling.
Top rake can be put on the teeth by milling it offset as shown in Fig.
88. Instead of the front of the tooth being milled on the centre line it is
RAKE ON CUTTER TEETH
115
cut off centre by distance x. Then ifR = radius of cutter and a the rake
x
angle required we see that — = sin a and x = R sin a .
R
Rakes up to 10° are advantageous, but above that angle they may cause
the cutter to chatter.
Blank
being Fluted
Fig. 88
Fig. 89
Example 1 . Calculate the amount of offset to give 10° of rake on the tooth
of a cutter 80 mm diameter.
In this case R = 40 and since sin 10° = 01736
x = 40 sin 10° = 40 x 01736
= 6944 (say 7 mm)
The case for an insertedblade cutter is shown in Fig. 89, where ifR =
radius over blades, then
x = R sin a as before.
Angle of fluting cutter
For milling the flutes in cutters the problem arises of determining the
angle a of the fluting cutter to give the required depth (d) of the flute
(Fig. 90).
116 MILLING AND THE MILLING MACHINE
Fig. 90
In the following analysis any land on the tooth is neglected, A being
assumed as a sharp edge. Also, the point B is assumed as a sharp corner,
so that if there is any radius on the cutter, B is the point at which the
cutter edges CB and AB would intersect.
The variation due to the first assumption tends to cancel out variations
due to the second.
AD is drawn perpendicular to CBO, then
/\
AD = AO sin AOC
= AD sin rr [where N = No. of teeth in cutter]
and since AO = rad of cutter, R,
360
AD = R sin
#
Now
tan a =
R sin
360
AD AD AD " J1 " AT
DB ~ CB  CD " d  CD ~ d  CD
But CD = CO  DO = R  AO cos
1 — cos
_ _ 360 J. 360\
R — R cos —fi = R II — cos —fi ]
360
360 '
N
R sin
Hence tan a =
360
</*(. cos**)
CUTTER CLEARANCE 117
Example 2. Calculate the angle of fluting cutter required to mill 16 teeth,
6 mm deep in a cutter 80 mm diameter.
Here R = 40, N = 16 and d = 6
R sin
tan a =
360
N
40 sin 22^°
6 40(1
1531
cos 22i°)
1531
6
79°6'
3044 2956
= 518
360
N
360
16
= 22^
From which a
The nearest cutter to use would probably then be an 80° cutter.
Clearance
The cutting clearance is put on the teeth at the time they are sharpened.
The cutting edges may be ground either on the periphery of a disc wheel
as shown at Fig. 91(a) or on the face of a cup wheel as at (b).
When the teeth are ground on the periphery of a disc wheel as at (a),
the centre of the cutter is set below the centre of the wheel and the radial
Grinding
Wheel
Fig. 91
118 MILLING AND THE MILLING MACHINE
line joining the tooth edge to the cutter centre is horizontal. The clearance
angle obtained is shown as C, and in triangle OBA:
AB . „
OB = SmC
But OB = radius of wheel (R)
and AB = offset (h c )
Hence ^ = sin C
h c = R sin C
Using the face of a cup wheel as at (b), the position of the cutter centre
relative to the wheel is immaterial, but the tooth is set below the cutter
centre by the distance h t and in triangle OED:
ED = h t ; OD = rad of cutter (/•)
ED . _ h t . _
t^ft = sin C or — = sin C
OD r
and h, = r sin C
Example 3. Calculate the settings for grinding the teeth of an 80 mm spiral
mill, (a) Using the periphery of a 200mm disc wheel and (b) using the face
of a cup wheel. (Clearance required = 6°.)
The first condition is as shown at Fig. 91(a)
and h c = R sin 6°
R = rad of wheel = 100mm
:.h c = 100 sin 6° = 100 x 01045 = 1045 mm
Using the cup wheel we have as in Fig. 91 (b)
h t = r sin 6° = 40 x 01045 = 418mm
Effect of the helical tooth on the clearance
When the tooth of the cutter is helical (spiral) the calculation for the
above cutter setting is modified as follows:
In Fig. 92 AK is perpendicular to the end of the cutter (parallel to its
axis), so that /$ is the helix angle of the tooth. The clearance face is AC and
/\
BAC = C a is the clearance angle referred to the end of the cutter, or to a
plane perpendicular to the cutter axis.
EFFECT OF THE HELICAL TOOTH ON THE CLEARANCE 1 19
Fig. 92
DG is parallel to AB and DF to AC, so that FDG = C a as before.
DL is horizontal and perpendicular to the tooth face AD, and DE is also
perpendicular to AD, but E is a point on the bottom of the clearance face.
G and H are vertically above F and E respectively.
y\
Hence EDH is the clearance on the tooth referred to a plane perpen
dicular to its front. Call this normal clearance (C„).
The axial clearance (C a ) is that put on by the grinding wheel, whilst the
normal clearance (C„) is the one effective when the cutter is in action. It
is necessary, therefore, to grind such an axial clearance as will give us the
required normal clearance, and for this purpose we require a relation
between C a and C n . This may be obtained by considering triangles EHD,
HDG and FGD, which are rightangled at H, H and G respectively.
tan C„ =
FG EH
DG ~ DG
(since EH = FG)
But
DH
DG
Hence from above: tan C,
— cos /J and DG
EH
DG
DH
cos/J
EH EH '
DH_=DH COS/}
cos fl
120 MILLING AND THE MILLING MACHINE
But — = tan C n
Hence tan C a = tan C„ cos /$
Thus when we know the value required for the normal clearance (C„)
and the helix angle /}, we can find the axial clearance angle C a for the
grinding setting.
Example 4. If in the last Example the normal clearance required was 6°
and the cutter had a helix angle of 30°, calculate the axial clearance
for setting.
We have that tan C a = tan C n cos /$
C„ = 6° and tan C n = 01051
fl = 30° and cos = 0866
tan C a = 01051 x 0866 = 0091
From which C a = 5° 12'
For small helix angles the correction is not important, but it should
be carried out on cutters with steep angles.
With the machine relieved cutter the clearance is put on at the time
the tooth is form relieved and the tooth is sharpened by grinding its front
with a saucershaped wheel. When this is carried out care should be
exercised to ensure that if the front of the tooth was originally radial (on
the centre), this position is preserved for it. If this is not done, the effect
on the formed profile will be similar to that we discussed when considering
the circular form tools on page 73, and the accuracy of the cutter form
will be lost.
Exercises 5a
1. Calculate the following for an 80 mm diameter fluted cutter:
(a) a suitable number of teeth to give a coarse pitch,
(b) the offset of the tooth front necessary to give 10° of front cutting rake,
(c) the setting for grinding the teeth on a cup wheel, to give 7° of clearance.
2. 300mm face mill is to be fitted with blades 6mm thick. Estimate a suitable number
of blades, and calculate the offset of the blade necessary to effect 12° of top cutting rake.
3. The flutes of a 25 mm end mill are cut LH helix 500mm lead. Assuming the flutes to
be 4 mm deep, calculate the helix angle, based on the mean diameter of the flutes. If the
end teeth follow this angle, is the rake on them positive or negative?
4. A spiral cutter is 60mm diameter and the teeth are cut on a helix of 300mm lead.
If an apparent clearance of 8° is ground on these flutes, what is the true clearance?
SPEEDS AND FEEDS FOR MILLING CUTTERS 121
5. Calculate the nearest angle of cutter to use for fluting a 60 mm diameter cutter
with 12 flutes, 6mm deep.
6. What depth of flute will be obtained by cutting 18 teeth in a 80 mm cutter with an
80° fluting cutter?
Speeds and feeds for milling cutters
The cutting speed for milling is found in the same way as for turning and
drilling, so that if Q mm = diameter of cutter, and N rev/min its speed,
Cutting speed (m/min)
TtDN , .. 10005
= T006 andAr = ^zJ
Cutting speeds should be as high as possible consistent with an econo
mic cutter life before it needs regrinding, and the speeds given for
turning form a reasonable basis upon which to set the speed of a milling
cutter.
The rate at which the work feeds beneath a milling cutter is sometimes
expressed in millimetres per minute and sometimes in millimetres per
revolution of the cutter. Neither of these methods gives a reliable indica
tion of the cutter performance since both ignore the number of teeth in
the cutter. The most equitable method of assessing milling feeds is in
millimetres per tooth, since this gives an indication of the work each tooth
is doing. From the feed per tooth, feed per revolution can be found by
multiplying by the number of teeth, and a further multiplication by the
rev/min gives feed per minute. The following table gives an indication of
the feed per tooth possible with various types of milling cutters:
Table giving feeds for H.S.S. cutters.
Cutter
Feed per Tooth
mm
Spiral (slab) mill (up to 30° helix angle of tooth) 01 to 025
Spiral mill (30°60° helix angle) 005 to 02
Face mill and shell end mill 01 to 05
End mill 01 to 025
Saw 005 to 01
Slotting cutter 005 to 01 5
Form cutters 005 to 02
Example 5. Calculate a suitable speed and feed for a 80 mm spiral mill
with 18 teeth to take roughing cuts on mild steel.
122 MILLING AND THE MILLING MACHINE
For roughing cuts we may take a moderately slow speed with a heavy
feed.
Assume a speed of 20m/min and a feed of 02 mm per tooth.
A , 10005 1000 x 22 x 7 __ , .
N = nD~ = 80x22 = 87 ' 5rev / min
A feed of 02 mm per tooth gives 02 x 18 = 36 mm per rev.
= 36 X 875  315 mm/min.
Power absorbed in milling
The conditions in milling are so variable that it is possible to attempt only
a very rough computation of power requirements. From experimental
work it has been found that the power requirements for milling are
approximately as given in the table below. The figures should betaken
only as a rough guide since the power varies with the amount of cut, the
cutter, the cutting lubricant and other factors. However, even if the
reader's, results are not all they might be, the working out will provide
useful mathematical practice.
ENERGY REQUIRED FOR MILLING*
Values given are Joules per cubic millimetre removed
Cast Mild Hard Alumin
Material being Cut ^ ^ ^ ^
mm
J/mm 3 19 27 40 to 70 160 090
*For face milling the power may be taken as f to \ of that given in table.
Using the values given in the table, the energy being absorbed is found
by multiplying the tabulated energy by the volume of metal being re
moved per minute. This is found by multiplying the depth of cut, the
width of cut and the feed length.
Thus if d = depth of cut; w = width of cut,
and / = feed
Volume = d.w.f.,
and Power = energy per second (watts)
To allow for frictional losses in the machine add approximately 30% .
THE DIVIDING HEAD 123
Example 6. A spiral milling cutter is taking a cut 4 mm deep with a feed
of 120mm per min over a castiron block 80mm wide. Estimate the
power required to drive the machine.
Here the depth of cut (d) = 4 mm
width of cut (w) = 80 mm
feed = 120mm/min = 2mm/s
Volume of metal removed per second = 4 x 80 + 2 = 640 mm 3
From the table we have l9J/mm 3 for cast iron.
.. Power for cutting = 19 x 640J/s = 1216W = l216kW
Adding 30% for machine losses we have
1216 X jjjjj = lS8kW
Exercises 5b
1. A 100 mm diameter spiral cutter has 18 teeth. Calculate the speed in rev/min and
the feed in mm/min for this cutter to be operating at a cutting speed of 22m/min and
a feed of 0 1 5 mm per tooth per rev.
2. If a cutter in Question 1 was operating on a job 80mm wide, with a cut 4 mm deep,
calculate the volume of metal removed per minute, and estimate the power input to the
machine if the material being cut is cast iron and frictional losses are equivalent to
30% of the cutting power.
3. If a milling machine is equipped with a 4kW motor, estimate the deepest cut that
may safely be taken on hard steel, when the work is 100 mm wide and the feed = 150 mm
per min. (Take cutting power as 75% of motor rating.)
4. A face milling cutter is 300 mm diameter and has 28 teeth. If this is operating at a
cutting speed of 33m/ min, and a feed of 0.2mm per tooth, how long will it take to travel over
a cut 400 mm long?
5. A certain job can be milled either with a 80 mm spiral mill, or with a 150 mm face
mill. Each cutter has 16 teeth and the cutting speed to be employed is 22m per min. If
the feed for the spiral mill is 02 mm per tooth and for the face mill 025 mm per tooth,
which is the most economical method of working?
6. A 25mm end mill with 8 teeth is milling a slot 10mm deep at a feed of0025mm
per tooth, and a speed of 600 rev/min. If the material being cut is brass, estimate the
power input (assume 30% loss) and the time and power cost for milling a slot 300 mm long.
Take power at 2p per B.O.T. Unit [1000 watt hours] .
The dividing head
The dividing head, which is shown diagrammatically'in Fig. 93 is used
with the milling machine for the purpose of obtaining divisions of the
124 MULING AND THE MILLING MACHINE
40 T. Worm
wheel
End for
attachmem
of work
Taper
hole Single start
worm
Plunger for locking
index plate
Crank Fig. 93
circle. It consists essentially of the spindle to which is attached a 40tooth
wormwheel. Meshing with this is a singlethreaded worm, to the spindle
of which is attached the indexing crank. Adjacent to the indexing crank
is the index plate containing several series of equally spaced holes
arranged in circles on its face. A pin in the indexing crank can be adjusted
so that its radius coincides with any of the hole circles, and an adjustable
sector enables any proportion of the index plate circumference to be
divided off.
Since the gear ratio in the head is 401, 40 turns of the crank cause the
spindle (and the work attached to it) to make 1 turn, or 1 turn of the
crank rotates the spindle ^th of a turn. The object of the index plate
with its holes is to subdivide further the turn of the crank, and the greater
the range of hole circles available the greater will be the number of
divisions possible without resource to special indexing methods.
The Brown and Sharpe dividing head is provided with three indexing
plates having hole circles as follows:
Plate No. 1: 15, 16, 17, 18, 19, 20 holes.
Plate No. 2: 21, 23, 27, 29, 31, 33 holes.
Plate No. 3: 37, 39, 41, 43, 47, 49 holes.
The standard Cincinnati dividing plate is of larger diameter than those
used on the Brown and Sharpe head and is reversible. It is provided
with the following hole circles:
On one side: 24, 25, 28, 30, 34, 37, 38, 39, 41, 42, 43 holes.
On the reverse side: 46, 47, 49, 51, 53, 54, 57, 58, 59, 62, 66 holes.
SIMPLE INDEXING 125
This plate enables all divisions up to 60 to be obtained in addition to
all even numbers, and numbers divisible by 5, up to 120.
In addition to this standard plate, special ones can be obtained when
divisions beyond its range have to be indexed.
Simple indexing
The majority of divisions required can be obtained without difficulty
by indexing in one of the sets of hole circles supplied. Straightforward
working in this way is usually termed simple indexing.
Since 40 turns of the crank cause 1 turn of the work, if we require
n equal divisions in the work each division will be th of its circum
n
40
ference and the turns required of the crank will be — . It will help the
reader to remember which way up this fraction should be if he remembers
that when more than 40 divisions are required on the work the crank
will have to rotate less than one complete turn.
Example 7. Calculate suitable indexing for the following numbers of
divisions: (a) 6, (b) 10, (c) 15, (d) 22, (e) 28, (J) 37, (g) 48, (h) 62.
(a) 6 divisions.
Indexing = f = 6£ = 6f = 6# or 6JJ
= 6 whole turns + 14 holes in a 21 circle
or 16 holes in a 24 circle
or any combination giving f of a turn
(b) 10 divisions.
Indexing = fg = 4 complete turns
(c) 15 divisions
Indexing = $ = 2f& = 2f
= 2 whole turns + f of a turn [see (a) above]
(d) 22 divisions.
Indexing = f§ = ljf = lft = lfi or Iff
= 1 whole turn + 27 holes in a 33 circle
or 54 holes in a 66 circle
(e) 28 divisions.
Indexing = ff = ltf = If = 1£ or lf
= 1 whole turn + 9 holes in a 21 circle
or 18 holes in a 42 circle
126 MILLING AND THE MILLING MACHINE
if) 37 divisions.
Indexing = $ = 1£
= 1 whole turn + 3 holes in a 37 circle
(g) 48 divisions.
Indexing — iQ — i — il— 20
lllUCAUlg — 48 — 6 — lg — 24
= 15 holes in an 18 circle or 20 holes in a 24 circle
(h) 62 divisions.
Indexing = f§ = §?
= 40 holes in a 62 circle or 20 holes in a 3 1 circle
Compound indexing
When a division is required which is beyond the capacity of the available
hole circles, a method of compound indexing may be used.
The index plate is usually locked and prevented from turning by means
of a plunger fitting in one of the circles of holes. The principle of com
pound indexing is to obtain the required division in two stages.
(1) By a movement with the crank in the usual way.
(2) By adding or subtracting a further movement by rotating the index
plate and controlled by the plate locking plunger.
Suppose the crank is indexed 5 holes in a 20hole circle and then the
index plate, together with the crank, is indexed a further hole with the
locking plunger registering in a 15hole circle.
If both movements have been made in the same direction the total
indexing will have been
to + rV = & + & = m on the worm.
If the plate had been turned opposite to the crank we should have had
J L — 11 _ J_ —  li
20 15 — ' 60 60 — 60
By compounding suitable hole circles in this way it is possible to obtain
a large number of additional divisions.
40
If n is the number of divisions required on the work, then — is
n
the indexing required and the fractions representing the two movements
40
to be used must give — , either when added or subtracted. Also the
n
denominators of the two fractions must be numbers equal to available
hole circles in the plate. Suitable hole circles must generally be deter
mined by a method of trial and error and the following examples will
illustrate the method.
COMPOUND INDEXING 127
Example 8 . Determine suitable compound indexing for the following divi
sions: (a) 77, (b) 91.
(a) 11 divisions.
The indexing required is $ and we require two suitable fractions which
give this when added or subtracted.
The method of trial and error may be assisted by the following work
ing.
Put down the 77 above a line, the 40 below it, and factorise them.
77 = 11 X 7
40 = 2x2x2x5
The numbers representing hole circles are now required to be written
below the 40 and factorised. Their difference, also factorised, must be
written above the line. The numbers must be so chosen that all the factors
above the line must cancel out with numbers below. [The reader should
notice that since only one plate can be used, the numbers must be those of
two hole circles on the same plate.]
Choosing 21 and 33 as the numbers, we have 21 = 7 x 3; 33 = 3 x 11
and the difference 12 = 2 x 2 x 3.
Putting these numbers with their factors down we find that all the
numbers above the line will cancel thus
77 = 44 X 7
12 = 2 x 2 x 3
40=2x2x2x5
21 = 1 x 3
33 = 3 x 44
Hence 2 1 and 33 circles will be suitable and we require to find the res
pective number of holes to be indexed. Let these be a and b.
tu a ^ b 40
Then 21 ± 33 77
(7 X 3) (3 x 11) (11 X 7)
Putting on a common denominator
11a ±lb = 3 x 40
7x3x11
i.e. 11a ±lb = 120
128 MILLING AND THE MILLING MACHINE
By trial and error we find that if a = 9, and b = 3; 99 + 21 = 120.
Hence the indexing required is 9 holes in a 21 circle added to 3 holes
in a 33 circle (i.e. both movements in the same direction).
(Jb) 91 divisions:
Testing for suitable hole circles as before we have:
91 =
44 X ^
10 =
4 x 2
5
40 =
2 x 2 x
2 X
39 =
44 x 3
49 =
7 x 2
49
as suitable sizes
and
a
b
40
39
± 49
=
91
(13
x 3)
(7X7)
(13 x
7)
Putting on a common denominator
49a ± 39b = 40 X 21
13 x 7 x 7 x 3
i.e. 49a ± 39b = 840.
By trial and error:
If a = 6 and b = 14:
49 X 6 + 39 x 14
= 294 + 456 = 840.
Hence 6 holes in a 39 circle added to 14 holes in 49 circle will be the
indexing required.
Differential indexing
This is really an automatic method of carrying out compound indexing.
The arrangement of the dividing head is shown in Fig. 94, from which it
will be seen that the index plate is unlocked, and is geared back to the
spindle. As the spindle is rotated via the crank and worm, the gear train
causes the index plate to turn backwards or forwards, and the net result
is the same as if the index plate were released and rotated by hand as in
compound indexing.
Differential indexing is more straightforward and is capable of dealing
with a wider range of divisions than compound indexing. The problem is
DIFFERENTIAL INDEXING
129
Index p/ate
'unlocked
Gear fixed to
index p/ate
Worm shaft
turned by crank
Gear fixed to
bevel wheel
Bevel wheels and
gears A+Bgive
J/7 ratio
Driver
Gear stud fixei
to head spindle
Fig. 94 Showing Arrangement of Gearing for Differential Indexing.
to calculate the indexing and the gear ratio necessary to obtain any given
number of divisions on the work, and we will explain the method by work
ing out a few examples:
Example 9. Calculate the differential indexing to give 107 divisions on
the work.
The indexing required is ^ turns of the crank per division, and when
this indexing has been done 107 times the crank has turned ^ X 107 = 40
times, i.e. the work has turned 1 complete circle as it should.
Since no 107 circle is available, let us take an approximately near
indexing to the exact ^ required.
ToV = i\ approximately (by approx cancellation by 5)
If we take 107 moves of 8 holes in a 21 circle we obtain
107 x 8
21
_ 416
~~ 21
40^
turns of the crank
130 MILLING AND THE MILLING MACHINE
But we have seen that the crank must make 40 turns only, during the
107 indexings, and we must therefore subtract jf of a turn. This is done
by gearing up the plate, so that whilst the spindle makes 1 turn, the plate
makes If turn in the opposite direction to the crank.
Hence with an indexing of 8 holes in a 21 circle, the gear ratio from
. ,, , , x Drivers 16
spindle to plate: =^ = ^rr
^ ^ Driven 21
For the Brown and Sharpe dividing head, the gears supplied are as
follows:
24(2), 28, 32, 40, 44, 48, 56, 64, 72, 86 and 100 teeth.
We may make up a train for the above ratio from this set as follows:
16 8 X 2 _ 64 32 Drivers
27 ~ 7~x~3 ~ 56 X 48 Driven
In the above case the gears must be arranged so that the index plate
revolves opposite to the crank.
Example 10. Calculate the differential indexing for 127 divisions.
Exact indexing = $;
and roughly cancelling by 8 gives ^ as the approximation.
Now 127 moves of 5 holes in a 16 circle gives 127 x & = W = 39f^ turns
of the crank.
But this is 4039{£ = re short of what it should be.
Hence in 1 turn of the spindle the index plate must make ^ turn in the
same direction as the crank.
4 . Drivers 5 5 x 1 40 24
Gear ratio: t=^ = tz = o » *•> = Ta x 7q
Driven 16 8x2 64 48
With this ratio and an indexing of 5 holes in a 16 circle, the 127 divisions
would be obtained.
Angular indexing
Very often, instead of a number of equal divisions, an angle must be
indexed .
Since 1 turn ofthe crank rotates the spindle^ turn, the angle at the work
360
centre equivalent to one turn of the crank is ^ = 9° so that
ANGULAR INDEXING 131
rx, i Angle required
Turns of crank to give any angle = ~
Example 11. Calculate the indexing for the following angles: (a) 41°, (b)
15° 30', (c) 29° 20'.
(a) 41°. Indexing = ^ = 4 turn of crank, say 4 whole turns and 30 holes
in a 54 circle.
(b) 15° 30'. Indexing = !£ = 1* = ift
= 1 whole turn and 1 3 holes in an 18 circle
or 39 holes in an 54 circle
id) 29° 20'. Indexing = th. = i±L = 3^ = 3£
= 3 complete turns and 7 holes in a 27 circle
or 14 holes in a 54 circle
Exercises 5c
1. Calculate suitable indexing to obtain the following divisions on the Brown and Sharpe
head:
(a) 12, (b) 15, (c) 22, (d) 34, {e) 41, (/") 50, (#) 62, (h) 76 divisions.
2. Calculate suitable indexing on the Cincinnati head for the following:
(a) 13, (b) 17, (c) 25, (</) 36, (e) 45, (/) 54, (g) 65, (A) 82 divisions.
3. Find suitable indexing for the following angles on the Brown and Sharpe head:
(a) 15°, (b) 26°, (c) 33° id) 52° 30', (e) 63° 40'.
4. Determine appropriate indexing for the following angles on the Cincinnati head:
(a) 16°30', (b) 27°45', (c) 31° 20', (^ 74° 15', (e) 136° 30'.
5. A shaft, 50 mm diameter, is to have a groove milled along it. The sides of the groove
are radial, it is 1 125 mm wide at the top and 6mm at the bottom. The centre is to be cut
with a cutter 6 mm wide, after which the shaft is to be indexed round and set over for milling
the slot sides with the same cutter setting. Calculate the indexing and set over.
6. Determine suitable compound indexing for the following, using B. & S. plates:
(a) 51, (b) 63, (c) 87, (d) 189 divisions.
7. The crank of a dividing head is indexed N holes in a C circle, and then the plate is
indexed in the opposite direction « holes in ac circle. Find an expression for the number of
divisions obtained.
8. Determine suitable indexing and gears for obtaining the following by differential
indexing:
(a) 97, (b) 53, (c) 101, (d) 131 divisions. [Use B. & S. plates and gears.]
. , . „,..,.,.. . . ,, . , . turns of spindle
9. The index plate of a dividing head is geared to the spindle in the ratio — . . — =
2
j (rotation opposite to crank) . If the crank is now indexed 3 complete turns, + 15 holes in a 20
circle, through what angle has the spindle been rotated?
132 MILLING AND THE MILLING MACHINE
10. A round plate requires four notches, A, B, C and D, indexing in it spaced so that the
angles between AB, BC, CD and DA are in the ratio 2:3:4:5. Using the same hole circle
throughout, determine the indexing required.
Use of continued fractions for angular indexing
When an angle is required whose value involves obscure minutes and
seconds, it is unlikely that an exact indexing for it will be possible. It is
possible, however, by turning the ratio into a continued fraction, to
obtain an indexing which is very near to the exact one. The method is
illustrated by the following examples:
Example 12. Calculate the nearest indexing and the actual angle obtained
for the following:
(a) 14° 38', (6) 21° 19' 35"
(a) The indexing will be — g — = 1— ^ —
Converting the fraction to minutes gives lffjj = l^fg. We now convert
f to a continued fraction and find its convergents.
169)270(1
169
101)169(1
101
68)101(1
68
33)68(2
66
2)33(16
32
1)2(2
The fraction is
1 +1
1 + I
1 + 1
2 + 1
16 +
USE OF CONTINUED FRACTIONS FOR ANGULAR INDEXING 133
The convergents are as follows:
1st = f; 2nd = ±; 3rd = f; 4th = f; 5th = $■; 6th = \
69
270
The 4th convergent is the last one we are able to make use of and
i = 10 flr 15
8 16 Ul 24
Hence the indexing is 1 complete turn and 10 holes in a 16 circle or 15
holes in a 24 circle.
The actual angle obtained will be
9 x If = 14f° = 14° 37*' (an error of *')•
21° 19' 35" 3° 19' 35'
(b) The indexing in this case will be s = 2 s and con
verting the fraction to seconds = 2£%$ = 2^
We now convert the 4& to a continued fraction
479
T296"
479)1296(2
958
338)479(1
338
141)338(2
282
56)141(2
112
29)56(1
29
The fraction is
27)29(1
27
2)27(13
26
02(2
2+ J
1 + 1
2 + \
2+ I
1+J
1 + \
13 + *
134
MILLING AND THE MILLING MACHINE
and its convergents are 1st = \\ 2nd = \\ 3rd = f; 4th = ^; 5th = $;
6th — a 7th — ia± 8th — ^^~
Ulll — 46 , /III — 625 , OU1 — !296
If a 46hole circle is available the 6th convergent may be used and the
indexing will be 2 whole turns +17 holes in a 46 circle, giving an angle of
9° x 2# = 21±f° = 21° 19'33" (an error of 2").
If a 46hole circle cannot be used, then the 5th convergent must be in
dexed and this will be 2 whole turns +10 holes in a 27 circle.
This will give an angle of 9° x2ff= 21}° = 21° 20' (an error of + 25").
Spiral milling
For some reason unknown, the operation of milling a helix on a cylindrical
piece of work has come to be known as spiral milling. Actually a spiral is
a fiat curve shaped like a clock spring. However, so long as we know the
process and its principles, it matters little by what name we call it.
Index plate
unlocked^*. Driven
Shaft i
marked r 'C 'J
In Fig. 94 1>J
Idler
F.g. 95 End View of Dividing Head Geared to Leadscrew for Spiral Milling.
SPIRAL MILLING 135
When the machine is set up for spiral milling the worm spindle of the
dividing head is geared to the leadscrew of the machine table, so that
when the leadscrew is turned the worm is turned also. This rotates the
dividing head spindle, so that the longitudinal movement of the table is
accompanied by a rotation of the work. A sketch of the machine set up is
shown in Fig. 95.
The first calculation necessary for spiral milling is the gear ratio bet
ween the leadscrew and the dividing head wormshaft to give the required
lead of helix.
In order to do this we must first ascertain the "lead of the machine."
The reader will recollect that the lead of a screw is the distance the screw
advances along the cylinder whilst it makes one complete turn round it
(Fig . 78a) . The lead of the machine is the lead of the helix it would cut if the
table leadscrew were connected to the dividing head worm by a f gear
ratio.
When this is the case we know that to rotate the dividing head spindle
(and the work) through 1 revolution requires 40 turns of the worm. If the
gear ratio to the table leadscrew is f the leadscrew will have made 40 turns
also, and the table will have advanced 40 (pitch of leadscrew). This
distance will be the lead of the machine. On the majority of milling
machines having metric leadscrews the pitch is 5 mm, hence the lead of
these machines is 200 mm (on machines manufactured to inch dimensions
the leadscrew pitch is 025 in, hence the lead of these machines is 10
inches).
When the lead of the machine is known, the gear ratio to cut a helix of
any lead is given by the proportion:
Drivers Lead of machine
Ratio =
Driven Lead of helix to be cut
The hand of the helix (whether RH or LH) is controlled by the presence
or otherwise of an idler gear in the train.
Example 13. Calculate suitable trains of gears to cut helices of the follow
ing leads on a machine with a 5 mm leadscrew: (a) 120 mm lead, (b)
256mm lead, (c) 480mm, (d) 720mm lead. [Select from the B. & S.
set of gears given on p. 130.]
As the leadscrew is 5 mm pitch the lead of the machine will be 5 x 40 =
200 mm.
136 MILLING AND THE MILLING MACHINE
(a) 120 mm
Drivers
Ratio j^t = ff = fi and this is given by a 40 T driving a 24 T.
The train may be compounded as follows:
40 5 X 8 _ 40 64 Drivers
24 ~ 4 X 6 ~ 32 X 48 Driven
(b) 256mm lead.
. Drivers _ 200 _ 100 x 2 100 x 24
Driven ~ 256 ~ 64 x 4 ~ 64 x 48
(c) 480mm lead
Drivers 200 _ 40 _5_ 40 32
Driven ~ 480 ~ 48 X 10 ~ 48 X 64
(d) 720 mm lead
_200_5_40_40 32
Ratio  7 20 ~ 9 X 80 ~ 72 X 64
Helix angle
Before the helix can be cut it is necessary to set the cutter to the angle fol
lowed by the path of the curve. If this were not done a great deal of inter
ference would take place and the shape of the groove would be nothing
like the shape of the cutter working to produce it. Even with the cutter
set into the helix angle some interference generally takes place and the
shape of the groove varies from the profile of the cutter.
The helix angle is the angle a on Fig. 78 (b), and if the helix is developed
out into a triangle as shown, we see that
circumference of cylinder _
lead of helix
When the helix angle has been determined, the cutter head or the
machine table must be swung round so that the plane of the cutter lies at
this angle, relative to the work. Care should be exercised to swing in the
correct direction for RH or LH helix.
When grooves of appreciable depth are being cut, the circumference of
the cylinder passing through the bottoms of the grooves will be much less
than that of the cylinder in which they are being cut. This means that if
the calculation for the helix angle is based on the outside diameter of
the work, the inclination of the cutter will be correct at the top, but not
at the bottom of the grooves. The reverse will be the case if the bottom
HELIX ANGLE 137
diameter of the grooves is used when finding the cylinder circumference
in the above expression.
When in doubt about which diameter to take for the calculation the
reader is advised to take the mean between top and bottom of grooves.
It may be that for grooves of certain shapes the top, or the bottom, might
form the more suitable basis for the calculation, but only trial and exper
ience can decide on the best compromise.
Example 14. Calculate the gears and setting for milling LH spiral flutes in
a reamer 40mm diameter. Lead of helix = 800mm. Machine leadscrew
5mm pitch. Reamer flutes 8mm deep.
Lead of machine = 40 x 5 = 200 mm
Gear ratio, table to head: r—  = ^r^ = t=o x t = 7o x ZT
Driven 800 4 2 2 48 64
To calculate the helix angle we will take the diameter at the mean depth
of the flutes, i.e. 40 — 8 = 32 mm.
Circum of cylinder = 327rmm
Circum 32;r n «,«
tan a = — 1 —  r  = ^r = ^f = 01257
lead 800 25
From which a = 7° 10'
As the helix is LH the table of the machine must be swung with its LH
end away from the operator.
Example 1 5 . A spiral gear has a pitch diameter of 80 mm and a spiral angle
of 30°. Calculate the gears for milling the teeth in it.
In this case we have to determine the lead of the helix in order to be able
to solve the problem.
The pitch circum = 80 n = 25 13 mm
circum
Also tan a =
lead
, , circum 2513 2513 A ~ c „
.. lead = — = ~7^ =rrznHA = 4352 mm
tan a tan 30° 05774
• , _■ j j i. j 200 1000 125
Gear ratio, leadscrew to dividing head = tttj = yyfz = 972
138 MILLING AND THE MILLING MACHINE
As this is an awkward fraction we will convert it to a continued frac
tion
125)272(2
250
22)125(5
110
15)22(1
11
7)15(2
]4
1)7(7
7
The convergents are given by
\
2 + 1
5 + 1
1 + 1
2 + }
and their values are:
1st = fc 2nd = A; 3rd = &; 4th #; 5th = §#•
Taking the 3rd convergent and finding a suitable gear ratio we have
Drivers 6 30 J_ _ 30 40
Driven ~ 13 ~ 65 X I ~ 65 X 40
The actual lead obtained will be ^ x 200 mm
= 4333 mm instead of the required 4352 mm
Cam milliing on the dividing head
By employing a universal milling machine fitted with a swivelling vertical
head, constant rise cams may be cut on blanks held in the dividing head.
The setup is shown in Fig. 96, and the dividing head is geared to the table
lead screw in the same way as for spiral milling.
The principle of the operation is that as the table moves to the right the
axis of the end mill approaches nearer to the axis of the dividing head
spindle. If, at the same time, the dividing head is geared so that its spindle
rotates, the combination of the rotation and of the cutter approaching
CAM MILLING ON THE DIVIDING HEAD 139
^Vertical head
swung over
Shaft
marked C
in Fig. 94
Gear connection
similar to that
for spiral milling
(Fig.9S)
Table
lead screw
Fig. 96 Setup for Cam Milling.
nearer to the centre, causes a cam to be cut on the blank held in the head.
The rate at which the end mill approaches the dividing head axis is con
trolled by the angle a, and the reader will observe that if this is made zero
(dividing head spindle horizontal), then, instead of a cam, a circular disc
would be cut provided the end mill was long enough to do it. (It will be
observed that as cutting proceeds, the blank gradually moves down the
end mill so that a fairly long cutter is necessary.)
(As an alternative to the arrangement shown, the end mill axis may be
located below the cam axis. Then the table must move to the left and the
cam will move up the end mill as cutting proceeds.)
140 MILLING AND THE MILLING MACHINE
The expressions for calculating any particular case may be derived
as follows:
Let a == angle of inclination (Fig. 96)
(Note that both cutter and dividing heads must be inclined at a)
R = Gear ratio =p between leadscrew and dividing head
worm
p = pitch of table leadscrew
/ == lead of cam to be milled
(Radial drop in profile in 1 revolution)
Then, for 1 turn of the table leadscrew, the dividing head worm will
turn R times and the dividing head spindle, because of the 401 reduction,
D
will turn 2^ times
Hence, since in 1 turn of its leadscrew the table advances pmm we
have:
R_
Turns of work 40 R
Movement of table p 40/)
From which
Turns of work = r~ (movement of table)
For 1 turn of the work:
1 = Tpj— (movement of table)
and movement of table = —J (for 1 turn of work)
K
Now because of the inclination of the head and work, when the table
has moved the distance cb (Fig. 97) the centre of the end mill has approach
ed the centre of the work by the length ac.
[^Movement of Tafa/e> Fig.
97
CAM MILLING ON THE DIVIDING HEAD 141
Hence if cb represents the horizontal table movement for 1 turn of the
work, ac will represent the lead of cam cut and from the triangle:
ac I
r = sin a or — t. = sin a
cb table movement
Hence Table movement =
sin a
If we substitute this in the expression for the table movement above we
have:
Table movement = — =/, i.e.—: = =r
R sin a R
, „ 40p sin a
and R = — ^—
lip = 5 mm, as is usually the case, the expression reduces to
200 sin a
R =
I
This expression controls the variables R and a , and in using it a value
for one of them will have to be assumed before the other may be calculat
ed. If a first trial gives unsatisfactory conditions, then one of the values
may be changed to bring the other to reasonable dimensions.
Example 16. Calculate a suitable setting for milling a cam, the profile of
which falls 12 mm in 100° of its angle. Leadscrew of machine has a pitch of
5mm.
Here p = 5 mm, and if the cam profile falls 1 2 mm in 100° , its lead will be
12 x tqtt = 432mm
Let us assume that R = 3
,, „ 40o sin a ~ 200 sin a
Then R =—i : 3 = 432
3 x 432
0648 = sin a
This gives a = 40° 24'
Drivers 3
Hence a gear ratio =r~ = y and an angular setting of a = 40° 24' will
be suitable.
142 MILLING AND THE MILLING MACHINE
Exercises 5d
1. Calculate the nearest indexing for the following angles, using the standard Cincinnati
plate, and determine the actual angle obtained for the indexing used:
(a) 10° 36' 30", (b) 41° 24'20", (c) 75° 45'30" .
2. Determine the nearest indexing to divide an angle of 76° 30' into 7 parts. Use hole
circles available on B. & S. plates.
3. A plate is rectangular, 123 mm x 79 mm, with a hole in its centre. Ifthis is on a mandril
between dividing head centres, find the nearest indexing to rotate it through the angle con
tained by joining the end corners to the centre. [Angle contained by short side required.]
In the following exercises use the B. & S. gears (p. 130) and take the Lead of Machine as
200 mm.
4. Calculate suitable gear ratios to cut the following leads on a B. & S. machine: (a)
168 mm, (b) 288 mm, (c) 450 mm, (d) 640 mm.
5. A gear has a pitch diameter of 100mm and the lead of the spiral is 420mm. Determine
a suitable gear ratio and find the helix angle for setting.
6. Calculate the gear ratio and angular setting for milling a 4start thread of35 mm pitch,
on a worm of 100mm pitch diameter.
7. A spiral milling cutter is 80mm diameter and 120mm long. Flutes are to be milled on
it which make approx £ of their lead in the length of the cutter. Calculate suitable gears
and angular setting for milling the flutes.
8. Slots h aving a lead angle of 30° are to be milled in a cylinder 45 mm diameter. Determine
the gear ratio.
In the following examples take the table leadscrew pitch as 5mm.
9. Calculate the lead of a cam which rises 10 mm in 50° of revolution, and determine a
suitable setting and gear ratio for cutting it.
10. A cam which revolves at 2 rev/min has to move the roller in contact with it at a rate of
50mm/min. Determine its lead, and find a suitable gear ratio and setting to mill its profile.
11. A cam is heartshaped with uniform rise. The radius to the point of the heart is 60 mm,
and to the corner where the curved portions meet the radius is 30 mm. Determine the lead and
find a suitable gear ratio and setting to mill it.
The calculation of solid angles
Angular milling and shaping jobs often occur where a solid angle has to be
calculated, the value of the angle depending upon the angles between
other surfaces on the same work. Examples of such cases are likely to be
varied, but the following worked cases may convey to the reader how any
other example might be approached.
Example 17. Calculate the true angle between the sloping faces of the
block shown in Fig. 98.
If we project a section through the angle, the section plane being taken
at 90° to the sloping corner, we shall obtain the triangle ABD shown at the
bottom oJ the diagram. The true angle we require will then be ADB.
THE CALCULATION OF SOLID ANGLES 143
Fig. 98
Now DE = CE cos 30°
and since triangle ABC is right angled at C and has 45° angles at A andB
Hence
CE = ±AB = 5.
DE=5cos30° =433 mm
AE /\
In triangle ADE (sectional view): ^^ = tan ADE
5 ^N
= tan ADE = 1155
433
/\
From which ADE = 49° 7'
/\ /\
ADB which we require is twice ADE = 2 x 49° 7' = 98° 14'
Example 18. Calculate the angle between the base and the sloping face of
the block shown in Fig. 99.
The block is shown by full and plain dotted lines. Chain dotted lines are
constructional only, for the purpose of explanation.
If AB is perpendicular to AG and BE, and BC perpendicular to BE
(BC being a line on the base), then the angle we require is angle ABC.
FC is parallel to EB, and BD to FE
144 MILLING AND THE MILLING MACHINE
Then in triangle BDC:
BD = FE =
Fig. 99
10
10
5774
/\
tan 60° 1732
BDC = 45° so that BC = BD sin 45° = 5774 x 07071 = 4083
10
AC /\
Buto^ == tan of ABC and AC
/\ 10
.\ tan ABC =
4083
ABC = 67° 49
2453
Example 19. Calculate for the tool shown in Fig. 100.
(a) The true angle between the cutting edge AB and the line AC
(b) The inclination of the top cutting face, perpendicular to AB
(c) The true angle between the top cutting face and the front clear
ance face.
Lines FE on the cutting face and EG on the clearance face are perpen
dicular to the edge AB.
FA is on the cutting face and is perpendicular to AC
FH and EK are vertical lines and angles FHG and EKG = 90°
In Fig. 100 (b), EM and FS are horizontal, LM vertical and EL is per
pendicular to AC.
(a) True angle between AB and AC
EL
If we call this angle a, then ry = tan a
EM EM
Now ^rr = cos 25° and EL = ^r
EL cos 25°
THE CALCULATION OF SOLID ANGLES 145
Fig. 100
Also
and
and
Hence
Hence
AL = Projected length AM
EM EM
Projected AM AL
EM
= tan 30°
AL
tan 30°
EM
tana =
EL cos 25°
AL
EM
EM tan 30°
X
cos 25°
EM
tan 30°
tan 30° 05774
= 06371
cos 25° 09063
a = 32° 30
(b) Inclination of top face perpendicular to edge AB
This will be the inclination of line FE
146 MILLING AND THE MILLING MACHINE
Now since points A, F, E and L are all in the same plane, and
FAL = FEA = 90°; AFE = EAL = a = 32° 30'
FE = FA cos a and FA = OR = PQ = —^
cos 25° cos 25°
Hence
FF 21 — 21 cos 32° 30' 21x08434 ,__.
FE == c^25^ C0S a = ~^Si2P = 09063 = 19  54mm
The vertical distance of F below E = ES
== Distance of F below A  Distance of E below A
= PR  LM
= PQ tan 25°  EL sin 25° =21 tan 25°  EL sin 25°
== 21 x 04663  EL sin 25° = 979  EL sin 25°
But EL = EA sin a and EA = FE tan a
.'. EL = FE tan a sin a
Hence EL sin 25° = FE tan a sin a sin 25° which, since FE = 1954, and
a = 32° 30'
= 1954 x 06371 x 005373 x 04226 = 283
PR  LM = 979  283 = 696 = ES.
The inclination of the top cutting face in a plane perpendicular to
AB = Angle EFS
FS /\ 6.06
and H = sin EFS = ^  03563
FE 1954
y\
From which EFS = 20° 52'
(c) True angle between top face and front clearance face.
This will be FES  8°
= 90°  20° 52'  8° = 90  28° 52' = 61° 8'
Exercises 5e
1. Determine the angle, when measured perpendicular to the clearance face, of a tool for
cutting acme threads (29° on its top face). Clearance angle on tool = 15°.
2. The angle of the vee in the block at Fig. 101 is to measure 60° on the front face as
shown . Calculate the angle of the vee when measured along its slope (i.e . the angle to which
it would be milled) .
THE CALCULATION OF SOLID ANGLES 147
Fig. 101
Fig. 102
3. Find the true angle a between the sloping face and the base of the block shown in Fig.
102.
4. A square pyramid is 40mm high and has a base 30mm square. Calculate (a) the angle
at the apex between two opposite faces, (b) the angle between two adjoining faces as mea
sured perpendicular to the sloping edge bounding them.
5. A piece of sheet steel 240mm x 120mm is bent as in Fig. 103. Determine the base,
height and vertical angle of a triangular piece of material which will fit into the angle as
shown dotted.
Fig. 103
Fig. 104
6. In Fig. 104 calculate (a) the angle between the corner AB and the top face of the
block, and (b) the angle between the two sloping faces as measured perpendicular to AB.
Mechanical
principles  I
Vectorial representation
We are familiar with the numerical representation of quantities and with
the processes to which we have to submit our calculations to obtain the
desired result. In the study of some parts of mechanics it is an advantage
to be able to express certain quantities in the form of vectors. When
anything has amount and direction it can be represented by a vector, and
a vector is a line, the length of which represents the amount of the quantity
being represented, and whose direction indicates which way the quantity
is acting. Thus in Fig. 105, ab represent a vector 3j units long directed in an
Fig. 105
upward direction at 45° to the horizontal; cd represents a vector 5 units
long directed horizontally from left to right.
Addition and subtraction of vectors
When we add or subtract numerical quantities we merely add or subtract
the numerical amounts, the result being a numerical sum or difference,
as the case may be. When adding or subtracting vector quantities,
however, we have to take into account not only their numerical value but
also their direction. This is achieved if we observe the following rules:
Adding
To add two or more vectors, draw the first in the direction of its arrow,
continue the second one on the end of the first, the third on the end of the
ADDITION AND SUBTRACTION OF VECTORS 149
second, and so on. The sum of the vectors is the vector joining the begin
ning of the first to the end of the last in the series. The arrow on the sum
vector must be in the same general direction as those on the vectors being
added. This will be understood from the vectors shown added in Fig. 106.
Draw ab equal and parallel to vector A ; on the end b drawee equal and
parallel to B and from c draw cd equal and parallel to C. The sum of A,
B and C is the vector ad, and its arrow is as shown.
Fig. 106
The chief mistake made in adding vectors is to add the succeeding
vector to the beginning instead of to the end of the previous one. This can
be avoided if the reader observes the following rule: When adding vectors,
do not remove the pencil from the paper until the end of the last one is reached.
Subtracting
We know that A  Bis the same as A + (  B) . Hence to subtract a vector
B from another one A we may add —BtoA.A minus vector is a positive
one with the direction of its arrow reversed.
This if a > b is a vector representing + A, then b < c will
represent — A.
The following example will illustrate the subtraction of vectors:
Example 1 . Subtract a vector of 5 units horizontal L to R, from a vector of
4 units directed at 45° to the NE, and then add a vector of 6 units vertical
downwards.
The vectors are shown at Fig. 107 (a), and the problem is
B A + C
150 MECHANICAL PRINCIPLES — I
A=5
+c ,r
C=6'<
Fig. 107
It does not matter in which order we deal with the vectors, providing we
observe the rules for their addition or subtraction.
In this case we have to reverse the arrow of A and then add them all
together. This is shown at (b) and the result is the vector ad.
To convince the reader that the result is unaffected by the order in
which the vectors are drawn, the diagram has been redrawn at (c), where
the vectors have been taken it the order C B A instead of B A C, as at (ft).
The result is the same for each case, since eh and ad are equal in length
and direction.
Exercises 6a
[Where an angle is specified, it refers to the angle made with the horizontal or vertical
line drawn to the beginning end of the vector.]
1 . Draw vectors to represent the following: (a) 7 units, vertically upwards; (b) 85 units
horizontal L to R; (c) 3 units upwards at 30° to L of vertical; (d) 64 units downwards 15° to
R of vertical; (e) 72 units R to L at 30° above horizontal; (/) 5 units L to R at 45° below
horizontal.
2. Add together the following vectors:
(a) 6 units vertically upwards, to 4 units horizontal L to R.
(/?) 82 units horizontal R to L, to 71 units vertically downwards.
(<•) 45 units downwards, 10° to R of vertical, to 65 units vertically downwards.
(if) 72 units R to L 20° above horiz, to 65 units L to R, 45° below horiz.
(<?) 63 units vert upwards to 6 units horiz L to R to 6 units R to L, 30° below horiz.
APPLICATIONS OF VECTORS 151
(/) 5 units horiz L to R, to 5 units vert, downwards, to 8 units L to R, 30° above
horiz.
3. In Ex. No 2, subtract the second vector from the first in (a), (b), (c) and (d).
4. In Ex. No 2, (e) and if), subtract the third vector specified from the sum of the
first two.
5. A horizontal vector, ab, 86 units long, arrow L to R, represents the sum of two vectors,
ac and cb, where acb = 90°. If ac is L to R, 40° above the horizontal, find the values of ac
and cb.
6. A vertical vector, ab, 10 units long, arrow downwards, represents the difference of two
vectors (ac  cb). Vector ac is 8 units long, downwards, 30° to R of the vertical. Find the
value of vector cb .
7. A vector ab, 15 units long, L to R, 45° above the horizontal, represents the sum of two
vectors, ac and cb. The angle between ac and cb is 30°, and ac is upwards, 30° to R of the
vertical. Find the values of ac and cb.
8. When two vectors, ab and be, are added, the result is a horizontal vector 10 units
long, L to R. When be is subtracted from ab the result is an upward vector, 30° to the R of
the vertical, 10 units long. If the angle between ab and be is 90°, find their values.
Applications of vectors
We will now consider some of the examples which occur in practice,
requiring the use of vectors for their solution.
Forces
In order to specify a force completely we must know its amount, its line
of action and its direction. Force may be represented vectorially since the
length of the vector may represent the amount of the force, the inclination
of the vector may represent its line of action and the arrow head will show
the direction. When problems arise where a number of nonparallel forces
are acting at a point, the solution can be arrived at vectorially, since this
method takes into account the angular effect as well as the magnitude of
the forces. A simple example will probably make this clear. Suppose that
a pin is being driven into a hole by a force F, applied on an angle as shown
in Fig. 108 (a). We know from experience that if F were large enough it
would eventually drive the pin home, although it would not do so as well
as if it were acting vertically. We also know that if F were sloping too far
over towards the horizontal, it would bend the pin . We may say, therefore,
that F is equivalent to the combination of a vertical force and a horizontal
one, and Fig. 108(6) shows the forces acting on the pin: Q is the pressure
of the side of the hole and R is the resistance tending to prevent the pin
from entering the hole. By drawing a vector diagram we may find Q and F
152 MECHANICAL PRINCIPLES — I
Fig. 108
if we know the amount and direction of F, because we know that F — Q +
R. The diagram is shown in Fig. 108(c). The vector for F is drawn parallel
to F and equal to its amount to some scale. Since only 3 forces are acting,
the diagram is a triangle, and this is completed by making one side parallel
to Q and the other parallel to R, Q, being drawn from one end of F and/?
from the other./? represents the driving in effect, and Q and bending effect
ofF.
The reader will observe that here the arrows follow round the diagram,
whereas when we were discussing the addition of vectors they did not.
Here we are dealing with vectors representing a set of forces which are in
balance amongst themselves, whilst before we were finding a vector which
represented the resultant of a number of others. If F were representing the
resultant, or the net effect of Q and R, its arrow would point upwards, but
as it represents a force which balances the other two, then its direction
must be reversed. The only difference between the resultant of a number
of forces and the single force that will balance them is that the balancing
force is opposite in direction to the resultant.
Example 2 . A round bar of metal which exerts a vertically downward force
of 100N is resting in a 90° vee block. Find the load on the sides of the
block.
In all problems of this type, a clear conception of the forces acting
should be gained before attempting the vectorial solution. Also, in all
APPLICATIONS OF VECTORS 153
100N
Fig. 109
vectorial problems, the data or details must be drawn out to scale in order
that the vectors may be drawn parallel to the quantities they represent.
In this case the weight of the bar (100N) acting vertically downwards is
being balanced by the forces R t andR 2 exerted on the bar by the vee block.
As the vee block is symmetrical about the centre line, the forces R x andi? 2
will be equal. The bar, of course, exerts equal and opposite forces, R Y and
R 2 , on the block (Fig. 109a).
In drawing the vector diagram, first draw a vertical vector to represent
the weight of the bar acting downwards. From each end of this draw
vectors parallel to the balancing forces acting. The lengths of these will
represent the magnitude of the forces. This is shown in Fig. 109(b).
It will be seen that in the vector triangle abc, since each side of the
vee block slopes at 45°, the angles abc are each 45° .
Hence
be = /?, = ac = R 2
= ab sin 45° = 0707 ab
= 0707 x 100 = 707 N
Example 3. A casting whose weight exerts a downward force of 1000N is
slung by chains as shown in Fig. 1 10(a). Find (a) the tension in the chains,
(b) the least angle a between the chains and the casting if the tension in
the chains is not to exceed 1600N.
154 MECHANICAL PRINCIPLES — I
1000 i k
b (*)
Fig. 110
(a) The forces acting at the ring where the chains meet are (1) lifting
chain pulling upwards, (2) sling chains pulling in their directions. These
forces are shown by the arrows.
The diagram must be set out to scale so that vectors may be drawn
parallel to their forces; this has been done for Fig. 1 10(a) to save an addi
tional diagram.
We may now draw the vector diagram, and this is shown in Fig. 1 10(b).
ab is the vertical vector for the upward pull of the lifting chain, whilst ac
and be are drawn parallel to the sling chains.
Upon measurement, ac and be are found to have a length representing
700N, which is the force in the chains.
(b) If the tension in the chains is 1600N, the vector diagram will be as
shown in Fig. 1 10(c), and by measurement the angle a is found to be 19° .
The reader will observe that the smaller the value of a, the greater will
be the load on the sling chains.
Exercises 6b
1. The leadscrew of a lathe is threaded 5 mm lead, and is connected to the spindle by a
f gear ratio. The feed shaft is set to give a feed of 1 mm per rev to the crossslide. If the nut and
the crossslide feed (outwards) are engaged at the same time, determine the actual move
ment of the tool point.
CONDITIONS FOR EQUILIBRIUM 155
2. A fitter holds a chisel at an angle of 40° to the horizontal, and strikes it a 50N blow
with a hammer. Find the force tending to drive the chisel horizontally, and that tending
to drive it vertically into the metal.
3. A bar of steel, which exerts a vertically downward force of 300N, rests symmetrically
on a pair Of 90° veeblocks.
(a) Determine the reaction between the bar and the block at an area of contact
(b) If contact takes place on an area of size 50 mm x 008 mm, what is the contact
pressure in N/mm 2 ?
4. A wheel, 60mm diameter, rolls along a fiat surface at 006 m/s. Determine the actual
speed and direction of a point on its circumference, and level with its centre.
5. A bar of steel 2m long and 31 kgf (304 N) weight, is lifted by a chain attached to its
ends. The total length of the vee formed by the chain is 25 m. Determine the tension in the
chain.
6. A casting weighing 20 kgf (196 N) is suspended by a chain and is being pulled to one
side by another chain attached to it. If, when the first chain is inclined at 30° to the
vertical, the angle between the chains is 105°, find the tension in the chains.
7. A casting of weight equivalent to 2000N is raised by driving 4 wedges under it. If
the angle of each wedge is 25° and the weight is equally distributed between them, find
the force tending to push the wedge out, and the pressure perpendicular to the wedge
surface. [Neglect friction between the wedge and casting.]
Conditions for equilibrium
For a body to be in equilibrium under the action of 3 forces, the forces
acting upon it must satisfy one of the following conditions:
(a) They must be parallel, or
(b) If not parallel they must meet at a point, and their vectors, when
drawn, must form a closed figure.
As we shall be dealing with parallel forces later, we will consider case
(b).
It is not always obvious, upon the examination of a problem, in which
direction all the forces are acting. Generally the points which they are
being applied can be picked out easily, as can the directions of at least two
of them. When we know that to be in equilibrium, nonparallel forces must
meet at a point, we can generally find this point by using what information
we have, and employ it to help in the solution of the problem.
In this connexion we might remind the reader of the following:
(a) The weight of a body always acts vertically downwards through its
Centre of Gravity, and if the mass is m kilogrammes, the weight can
be taken as 981 m newtons.
(b) When friction is neglected the pressure between two bodies acts at
the point of contact and in a direction perpendicular to a tangent
drawn to the surfaces in contact.
156 MECHANICAL PRINCIPLES — I
Example 4. A bar of metal, mass 15 kg (147N weight), is rested on its end
and leaned against a wall so that it is inclined at an angle of 60° to the floor.
If the friction between the end of the bar and the wall is neglected, find the
force on the floor and against the wall.
The forces acting on the bar are (1) The gravitational pull of its weight,
acting downwards through its centre; (2) the reaction of the wall acting
horizontal (since friction is neglected); (3) the reaction of the floor. Of
these we know the amount, direction and line of action of ( 1 ), the direction
and line of action of (2), and the point of application of (3).
If lines representing ( 1 ) and (2) are drawn on the diagram they meet at O
[Fig. 1 1 1 (a)] . Obviously, if the forces acting must meet at a point, the third
force must pass through O, and since it must also act at the end B of the
bar, its line of action is OB. Having thus determined the directions of all
the forces acting, and knowing that they must form a closed vectorial
figure, we may now draw the vector diagram.
I
147 NT
Fig. Ill
Draw ab to represent the weight of the bar, be parallel to OB, and ca
parallel to AD.
From the diagram we find that reaction of floor (be) = 153N and load
on wall (ca) = 40N.
Example 5. A bell crank lever ABC is pivoted at B and the forces acting at
A and C are as shown in Fig. 1 12(a). Find the force on the pivot B.
FORCES ACTING ON A CUTTING TOOL 157
If the lines of action of the forces at A and C are continued they meet at
D [Fig. 112(6)], so that the third force acting on the lever must pass
through D. Since the lever is supported at B, this force must also pass
through B, so that the forces acting on the lever are as shown by the
arrows.
>A
200N
60
(«)
B
120
C
100Nrr
Fig. 112
The vector diagram of forces is shown at Fig. 1 12(c) and from it we see
that force acting at B = ac
= V200 2 +
= 224N
100 2
Forces acting on a cutting tool
We have noticed previously that when a lathe tool is cutting there are three
perpendicular forces acting on it. These are shown diagrammatically in
Fig. 1 13(a), where Cis the vertical cutting force, Fthe feeding force, and//
the horizontal pressure of the work. By means of vector diagrams we may
determine the resultant of these forces.
Since the forces are acting in two planes we must solve the problem in
two stages. Let us find the resultant of F and C first. This is shown by the
vector diagram abc at (b), and R FC , represented by ac, is the resultant, act
ing in the vertical plane containing Fand C. We may now imagine the tool
being acted upon by the forces R FC and //, acting in a plane inclined at to
158 MECHANICAL PRINCIPLES — I
Fig. 113
the vertical as shown at (a). Combining these in a second vector diagram,
we obtain the final resultant force on the tool (R) as shown at (c).R lies in a
plane sloping at 6 to the vertical and its line of action is inclined at a to
the line of H . A diagrammatic sketch of the forces and their resultants
is shown at (d).
The reader will observe that we might solve this problem by calculation
f
alone, since from Fig. 113(6) F 2 + C 2 = R 2 , and^ = tan 6; and from
C
(c)R 2 = R 2 + H 2 and tan a =^
H
In all such cases, however, it is advisable to be able to visualise the
conditions and not to calculate a result blindly from a formula.
Example 6 . The forces acting on a lathe tool are C = 1000N, F = 100N and
H = 400N [Fig. 113(a)] . Find the resultant force on the tool.
THE BALANCING OF WORK ON LATHE FACEPLATES 159
As we have discussed the problem diagrammatically, we will solve this
problem by calculation.
Referring to Fig. 113(6), (c) and (d).
R fc = F 2 + C 2 = 700 2 + 1000 2
R FC = V700 2 + 1000 2 =
. . F 700 n _
tan0 = C = TOOO = a7
from which 6 = 35°.
1220N
R* = R FC * + H 2
1220 2 + 400 2
R = V1220 2 + 400 2 = 1285N
tan a =
R.
1220
= 305
H 400
from which a = 71° 51'
Hence the resultant is a force of 1285N, and the angles a and 6 in Fig.
113(fi0 are 71° 51' and 35° repectively.
The balancing of work on lathe faceplates
When a mass is rotating at a certain distance from the centre of rotation
the disturbing effect due to its being out of balance is proportional to the
mass and to its distance from the centre of rotation. Furthermore, the
disturbing effect is always directed from the centre to the masses. Thus if
Fig. 1 14 represents a rotating plate carrying rotating masses m, and m 2 at
radii r l and r 2 , the outofbalance forces are proportional to m l r 1 andm 2 r 2
and are directed from the axis (O), through the mass centres.
We are thus able to deal with such problems vectorially, provided we
draw our vectors equal in length to the product mr.
Fig. 1 14
160
MECHANICAL PRINCIPLES — I
Example 7. A rotating plate carries masses of 15 kg and 20 kg placed at
radii 80 mm and 100 mm respectively. The angle between the masses is
120°.
Find where a 22 kg mass must be placed to balance the system.
The two masses are shown at Fig. 115(a).
Wrfor pl}~** _ .
<t of Balance
Weight
Fig. 115
The products of their mass and radii are
15 X 80 = 1200
20 x 100 = 2000
The vector diagram is now drawn with the vector lengths proportional
to 1200 and 2000, the directions of the vectors being parallel to radii join
ing the c entre O to each respective weight . This is shown at abc in Fig .115
(b) and ca is the vector representing the product mr for the balancing mass.
From the diagram the length of ca is 1700 units, and since for the balance
1700
weight m = 22 kg, r
22
= 77 mm.
The position of the balance weight relative to the other two is found by
drawing a line through O parallel to ca . This is shown dotted on Fig . 1 1 5(a) .
Hence to balance the masses given, the 22 kg mass must be fixed at 77 mm
radius, and 97° from the 15kg mass.
When working out the balancing of irregularly shaped castings, the
point or points must be determined where the mass of the casting is acting
(the Centre of Gravity). If the metal is all concentrated together, the
centre of gravity may be determined as a single point, but if the casting
consists of lumps of metal concentrated at different positions, a better
method would be to estimate them as separate units of mass.
APPLICATIONS OF VECTORS
161
Exercises 6c
1. The forces acting on a lathe tool are 1200N vertical, 800N along the axis of the work
and 500N outwards, perpendicular to the axis of the work. Find the amount and direction of
the resultant force.
2. If the tool in question 1 is 20mm deep, how far back from its point does the line of
action of the resultant force intersect its base?
3. The helix angle of the tooth of a spiral milling cutter is 20°. When the tangential
cutting force perpendicular to the cutter axis is 850N, find the end thrust and the force
acting perpendicular to the tooth face.
4. The arms of a bell crank lever are 60mm and 40mm long. When the long arm is
inclined at 45° to the upwards vertical, the short arm is below the centre line, and with the
lever in this position a vertical upwards force of 120N at the end of the long arm is balanced
by a horizontal force of 180N at the end of the short arm. Find the force acting on the
lever pivot.
5. A bar of steel weighing 31 kgf (304 N) rests with one end in the corner between a wall
and the floor and is inclined at 30° to the horizontal. It is held in this position by a rope
attached to the outer end, the rope making an angle of 90° with the bar. Find the tension in
the rope and the force where the bar rests in the corner.
6. A countershaft is driven by a horizontal belt and the down belt makes an angle of
75° with the top one. When the tension in each side of the driving belt is 200N and in each
side of the down belt 120N find the resultant force on the shaft.
350 N
rmXJ
r
y \Z50N
Fig. 116
7. Fig. 1 16 is a diagrammatic sketch of a gear drive, and forces of 350 N act on the teeth of
the 100mm gear as shown. Find the resultant thrust on the bearings of this gear.
8. A toggle press mechanism is as shown in Fig. 121. When the horizontal force at C is
250N and the angle CA 2 D is 5°, find the vertical force at the ram.
9. A casting is bolted to a lathe faceplate, the total mass being equivalent to
162 MECHANICAL PRINCIPLES — I
masses of 50kg at 200mm radius and 60kg at 260mm radius. The angle between these is
120°. Find the radius and position of a 50kg mass which will give balance.
10. For the mechanism shown at Fig. 126, if a load of lOkN acts vertically down
wards at E, what is the thrust in link BC?
11. A large CI pulley has two outofbalance bosses on it each 60 mm diameter and
40mm thick. Their radii are A = 260mm and B = 360mm, and the angle between radii
drawn to A and B is 110°. Find the volume and position of a lead balance weight to be
attached to the pulley at 360mm radius.
[Density CI = 75g/cm 3 , lead = ll3g/cm 3 ],
Vector diagrams of velocity
Velocity, as well as force, may be represented by a vector since it has
amount, line of action and direction. The application of vector velocity
diagrams is very useful when studying the speeds of points in machine
mechanisms, and for that reason it is worthy of consideration.
Before going further into the subject it will be well to impress certain
fundamental points on the reader's mind, as the success or otherwise of
his further study will depend upon his appreciation of them.
Fig. 117
(1) When an object is rotating in a circle its velocity at any instant is
directed perpendicular to the radius upon which it lies. If co rad/s is the
speed and r is the radius, then the velocity is cor. For most engineering
applications the rotational speed is usually quoted in rev/min and the
radius in millimetres. For general purposes the most convenient unit for
velocity is metres per second, and hence if
N = speed in rev/min
and r = radius in millimetres
,, 2nrN 7idN A . ,
then V = 60000 = 60000 metres / second
VECTOR DIAGRAMS OF VELOCITY 163
(2) Whatever may be the motion of a rigid rod, the only velocity that
one end may have relative to the other is perpendicular to the rod. (The
"relative" motion of one body to another is the motion the first would
appear to have to an observer situated at the second one.)
Fig. 118
In Fig. 1 18, AB is a rod having any motion whatsoever. To an observer
at A, B can only appear to move perpendicular to AB. If it could move
in any direction other than this, it must either approach nearer to A
or recede from it. Both of these are impossible since AB is rigid and of
fixed length.
(3) When a part of a machine is guided in slides it can only move parallel
to the slides.
We will discuss vector velocity diagrams by working one or two pro
blems, and the reader is advised to take particular note of the system of
lettering adopted.
Example 8. ABC is a slider crank mechanism. Find the speed of C for the
position and values given (Fig. 119).
Speed of B =
27irN 2 x 22 x 60 x 500
60000
= 3142m/s
7 x 60000
500 ,
rev/min
Fig. 1 19
164 MECHANICAL PRINCIPLES — I
From the diagram we know the following:
(1) B is moving perpendicular to AB at 3142m/s per sec,
(2) Relative to B, C can only move perpendicular to BC,
(3) Relative to A, C can only move horizontal.
306 c Fig. 120
The velocity diagram is shown at Fig. 120 and the explanation of its
construction is as follows:
(Arrows are shown on the vectors, but in practice these are omitted.)
ab represents the velocity of B relative to A (i.e. the velocity B would
appear to have to an observer at A). The relative velocity of C to B is
perpendicular to BC, hence from b a line is drawn perpendicular to BC,
and c must lie somewhere on that line. But relative to A, C can only move
horizontal, hence c must lie somewhere on a horizontal line through a.
The point c is therefore given by the intersection of the two lines drawn,
and abc is the velocity diagram. For the given position C is moving
towards A at a speed of 306 m/s, scaled off from ac.
[Note in the lettering of velocity diagrams that ab = velocity of B
relative to A and not of A relative to B as might have been expected.
Similarly for ac, be, etc.]
Example 9. The mechanism of a toggle press is shown in Fig. 121 (a).
Find the speed of the ram for the position shown. The figure is drawn
to scale.
When the frame (the fixed element) appears in more than one place it
should be given the same letter with small figures. In this case A, and
A 2 are Iboth fixed frame points and are lettered accordingly.
The vector velocity diagram is shown at Fig. 121 (b).
.. . . + e D InrN 2 X 22 X 60 X 50
Velocity of B = ^^ = 7x60000
= 0.314m/s
VECTOR DIAGRAMS OF VELOCITY 165
260
Fig. 121
Draw ab perpendicular to A,B to represent the velocity of B relative
to A, . From b draw a line perpendicular to BC, and from a draw per
pendicular to A 2 C. The intersection of these lines gives the point c.
From c draw a line perpendicular to CD and from a draw a vertical line
(velocity of D relative to A is fixed as vertically) . The intersection of these
gives point d.
The vector ad represents the relative velocity of D to A (the frame).
This is shown to be downwards and scales off from the diagram to be
0052 m/s. If the speed and direction of a point on any lever is required, it
can be found immediately from the vector diagram, since lines on the
vector diagram are images of corresponding ones on the mechanism.
Thus for a point E, § the distance from C to D, locate the point e,
§ from c to d, and ae represents the velocity of E.
There are one or two additional points to be explained in connection
with velocity diagrams, and these will be covered by the next example.
Example 10. The mechanism for a slottedlink shaping machine quick
return motion is shown drawn to scale in Fig. 122 (a). Determine the
speed of the ram for the position shown.
The crank A,B rotates, and the block B moves up and down in the
slotted lever A 2 D . This lever is thus made to oscillate about its pivot
A 2 2, and through the link DE drives the ram of the machine.
166
MECHANICAL P.
When dealing with a mechanism in which the end of one lever slides
in Or on another lever which itself moves, the lettering up should be
arranged as shown. B represents the block, and C represents a point on
the lever A 2 D. B is rotating about A, whilst C is rotating about A 2 . The
reason for doing this will be seen when we discuss the velocity diagram.
vi •♦ f D 2 x 22 x 80 x 40 .,__ .
Velocity of B = 7 x 600 oo = °^ 5 ^
Ram
'in andBJock)
C( Point on Lever
\ /ajd;
Fig. 122
In Fig. 122 (b) ab represents the velocity of B relative to A. Now the
velocity of C relative to B (or B to C) is along the lever A 2 D. From b
draw a line parallel to A 2 D and from a draw a line perpendicular to
A 2 D (since relative to A, C is moving perpendicular to A 2 D). The inter
section of these gives the point c. The line ac now represents the relative
velocity of C to A. The velocity of D relative to A will be a continuation
ac
oiac to d such that — j
ad
A 2 C
A,D
because C and D are on the same lever and
their movements are proportional to their respective radii. (For this
FORCES ACTING IN A MECHANISM 167
purpose C is assumed to be coincident with B, as it is only an imaginary
point introduced for this purpose of the construction.)
When d is thus located, draw a line through it perpendicular to DE,
to represent the velocity of E relative to D and draw a horizontal line
through a to represent the relative velocity of E to A. The intersection
of these gives the point e. From the lettering it will be seen that E is
moving to the right and the scaled length of ae gives its speed to be
0353 m/s. The speed at which the block B is sliding in lever A 2 D is given
by the length be on the velocity diagram.
Forces acting in a mechanism
The vector velocity diagram for a mechanism may be used to estimate
the force acting at a certain point when the force at another point is
known. Let us assume that for a mechanism we have the velocity of
two points A and E, and the force acting at A. If v A and v E are the
velocities, and F A and F E the forces, respectively, then assuming A to
be the energy input point and E the point of output, the input rate of
work at A will be F A v A and if the machine is 100% efficient, this will
equal the output rate F E v E .
We can assume an efficiency r] and then we shall have
_ Output _ F E v E
v ~ Input ~ F A v A
or, since we know F A , v A and v E ,
F _ t ?( F a v a)
E V E
Example 11. If the torque input to the toggle press in Example 9 is
lOONm and the efficiency of the mechanism is 60%, estimate the pressure
on the ram for the position shown.
Since torque = (Force)(radius), and r = 60 mm = 006 m
100 = (Force)(006) and the force at the end
100
of the crank = j^^ = 1670N
UUo
Here we have that v B = 03 14 m/s, v D = 0052 m/s, and F B = 1 670 N
Hence efficiencv  FpVp 06 = F d x °° 52
Hence emciency  — . uo im x Q314
From which F D = ' 6 X ^ ' 314 = 6000 N
168 MECHANICAL PRINCIPLES — I
Exercises 6d
1. In a slidercrank engine mechanism similar to Fig. 119 the crank AB is 60mm long
and the connectingrod BC is 220 mm. Find the speed of C when AB is rotating at
lOOOrev/min and (a) the angle BAC is 45°, (b) angle BAC = 120°.
2. For the problem in Question 1, find the velocity of the midpoint of BC, when
angle ABC = 90°.
3. Fig 123 shows a quick return shaping machine drive in which cutting takes place
when C moves to the left. Find (a) The ratio cuttl "g timc t b \ The speed of B in rev / min
return time
if C is to have a velocity of 024 m/s when at the midpoint of its cutting stroke.
Fig. 123
4. In Fig. 124 crank AB revolves at 200rev/min and CD is caused to rock backwards
and forwards. When AB is at 30° to the horizontal as shown, find the speed of C and
of E, the midpoint of BC.
5. In Fig. 125 A and B are two blocks which slide in slots at right angles. If AB = 240mm
and AC = 80 mm, find the speed and direction of C when angle OBA = 40° and A is
moving downwards at 024m/s.
Fig. 125
Fig. 126
APPLICATIONS OF VECTORS 169
6. In Fig. 126 crank AB rotates as shown and CDE is a solid bell crank lever. Find
the speed of E when AB is at 20° to the vertical.
7. In a toggle press mechanism similar to Fig. 121, A,B = 40mm, BC = 240mm,
A 2 C = CD = 160mm. The vertical centre lines are 280mm apart and the centre line of
A, is 120mm below the centre line of A 2 .
When B is rotating at 60rev/min and angle CA 2 D is 10°, find the speed of D.
8. In the last problem if the torque on A,B is 100 Nm, find the load at D if the
overall efficiency is 60%.
9. In the press mechanism shown in Fig. 127, find the speed of E when D is moving
upwards at 0.02m/s (A,E is horizontal). If in this position the load at D is 4000N,
what force can E exert if the overall efficiency is 50%? [CA,E is a solid lever.]
Path of C
A,B = 170 ~i^ \
A 2 C = 120 ^>
EC = 300
Fig. 127
Fig. 128
10. In the Whitworth quickreturn mechanism sketched in Fig. 128, B rotates about
A, and slides along the lever CD. CD is pivoted at A 2 .
Time of outer stroke, ^ , „w. jnu ad*
Find (a) the ratio for E, and (b) the speed of E when A,B is at
Time of inner stroke
45° as shown.
11. In Fig. 129 AB is a door hinged at A. CD is a springloaded arm for closing the
door and hinged at C. If B is moving at 036 m/s, find the speed at which D is sliding along
the door when the door has opened 45° .
Fig. 129
170
MECHANICAL PRINCIPLES — I
The moment or turning effect of a force
If a force acts on a body, and the effect of the force is considered relative
to some point not on its line of action, the tendency is for the force to
rotate the body about the point. This tendency is called the moment
of the force about the point. Thus in Fig. 130, if the effect of the force
F is considered relative to the point 0, F tends to rotate the body about
O . The numerical value of the moment of F about O is found by multiply
ing F by the perpendicular distance from O to its line of action. Thus the
moment of F about O = Fx. If F were rotated to act in some other direc
tion (e.g. as shown dotted), then its moment would h^F, (perpendicular
distance) = F,_y.
To calculate turning moment we multiply a force by a distance so
that the unit of turning moment is usually newton metres (Nm) (although
on occasions a unit such as Nmm may be more convenient to save
converting and reconverting units of length).
Fig. 130
Fig. 131
When two equal and opposite forces act on either side of a pivot they
constitute what is called a Couple, and the turning moment exerted by a
couple is given by one of the forces multiplied by the perpendicular
distance between their lines of action (Fig. 131).
Total moment of forces P = Fr + Fr
= 2Fr = F(2r)
= F(perp distance between forces).
A good example of a couple is given by the forces applied to a tap
wrench or die stocks when cutting a thread. The couple applied is equal
to the turning moment resistance at the cutting edges of the tap or die.
CONDITIONS FOR EQUILIBRIUM WITH MOMENTS ACTING 171
Moment of a force about a point on its line of action
Since moment = (Force)(distance), if the distance is zero, then the
moment will be zero if the point lies on the line of action of the force.
Thus in Fig. 130 the moment of F about A = F x O = 0.
Conditions for equilibrium with moments acting
The reader will have observed that a moment may be directed clockwise
or contra clockwise.
The condition for a body to be in equilibrium isthat the sum of the clock
wise moments about any point must be equal to the sum of the contraclockwise
moments. When this condition is satisfied there will be no unbalanced
turning effort causing the body to rotate.
Example 12. A shaft is 75 m long between the bearings and carries 3
pulleys spaced at 2 m, 3 m and 6 m from the LH bearing. The downward
force on the pulleys due to the belt drives is 300 N, 250 N and 350 N
respectively. Calculate the load on each bearing (Fig. 132).
75m
Fig. 132
Since the loading on the shaft by the pulleys is downwards the forces
exerted on the shaft by the bearings will be upwards. Call these R A
andi? B .
We thus have the shaft acted upon by a system of parallel forces, and
since it is in equilibrium, the conditions to be satisfied are:
(1) Total upward forces
(2) Clockwise moments
Total downward forces.
Contraclockwise moments.
If we take moments about the bearing A, we can neglect the force
acting there since it will have no moment about that point.
172 MECHANICAL PRINCIPLES — I
Hence by moments about A
Clockwise moments = 300 x 2 + 250 x 3 + 350 x 6
= 600 + 750 + 2100 = 3450Nm
Contraclockwise = R B X 75Nm
These must be equal
3450
75 R B = 3450 and R B =
460 N
Also, since upward forces = downward forces:
300 + 250 + 350 = 900N
*a + *b
R A = 900
R,
900  460 = 440 N
Example 13. A system of levers is shown in Fig. 133. Calculate what load
hung at W may be balanced by a force of ION at A. [Neglect friction
and the weight of the levers.]
R
H 20 h
T
\
^ir?
\
240
200
J$2
Wu
kip
\15
a
^
"10N
Fig. 133
First let us consider the top lever, which will be balanced by the force
F in the connecting link and the ION at A.
Taking moments about B, and working in units of N and mm, we
have
Clockwise = 10 x 240 = 2400 and contra clockwise = 20F
Hence 2400 = 20 F and F = 120N
This will pull down on the top, and upwards on the bottom lever.
Now consider the bottom lever and take moments about C.
Clockwise = F x 215
Contraclockwise = W x 15
EXAMPLES INVOLVING MOMENTS 173
These must be equal, and since F = 120
120 x 215 = \5W
W = 12 ° * 215 = 1720N
The actual value of W would be affected by friction, the weight of the
levers and of the stirrup. If the lever weight is appreciable it can be
allowed for by assuming the weight of each lever to be acting as a down
ward force approximately at the lever centre. For example, if each lever
weighs ION and the weight is assumed at the lever centre, we have for
the top lever:
20 x F = 10 x 240 + 10 x 110
= 2400 + 1100 = 2500
F = 175N
For the bottom lever
F x 215 = 15W  10 x 1075 and since F = 175N
175 x 215 = 15 W 1075,
15 W = 37 625 + 1075 = 38 700
yJg™2580N
To obtain an exact calculation, the centre of gravity of the levers could
be found by balancing on a knifeedge. The weight is then taken as
acting through the centre of gravity.
Example 14. If in Example 6, p. 158, the work is 240mm long between
centres, and the cutting tool is 60 mm from the tailstock centre, calculate
the forces acting on the centres.
The forces acting on the work are as follows:
(1) Vertical upwards force of 1000N.
(2) Horizontal force of 400 N directed away from the tool.
(3) Longitudinal dorce of 700 N towards the headstock.
These are shown in Fig. 134 (a).
Let us consider the vertical force first. This will cause an upward
force on each centre and an equal and opposite downward force on the
work by the centre.
Taking moments about centre A, and working on units of N and
mm we have
Clockwise moment = (vert force at B)240
174 MECHANICAL PRINCIPLES — I
1000N
Headstock
(a).
Fig. 134
(b).
Contraclockwise moment =1000 x 180 = 180 000
Hence 240 (vert force at B) = 180 000
180 000
Vert force at B =
Vert force at A =
240
1000
= 750 N
750 = 250N
In the same way we have that the 400N force is balanced by horizontal
forces of 300 N at B and 100N at A.
The 700 N longitudinal thrust is transmitted through the bar entirely
to A.
We thus have forces acting on the two centres as shown by Fig. 134(b),
and as an additional exercise the reader should combine these vectorially
to find the resultants.
Exercises 6e
1. A bell crank lever has equal arms of 120 mm at 90°. When the lever is pivoted with
one arm vertically upwards, a weight of 120N is hung on the end of the horizontal arm.
What force, inclined at 45°, must be applied to the end of the upper arm to balance the
lever?
2. A 20 mm hand reamer is being operated by hand pressure at each end of a lever
480mm long. The hole is 20mm long and each of the 6 teeth is taking a cut of 004 mm
If the cutting pressure at the teeth is 80N per mm 2 of cut, estimate what force must
be applied at each end of the 480mm wrench.
3. A bar is 240mm long between lathe centres. When the tool is 90mm from the tail
stock the vertical cutting pressure is 850N. Find the vertical force on each centre due
to this.
4. A piece of material is held in a lathe chuck and the point at which cutting takes
place overhangs from the centre of the front bearing by 125 mm. If the vertical pressure
due to the cut is 1200N, and the bearings are 360 mm apart, estimate the force on each
EXAMPLES INVOLVING MOMENTS 175
bearing. [Assume point contact at each bearing and neglect weight of spindle and chuck.]
5. The bearings of a countershaft are 900mm apart. The horizontal force due to
the top belt is 450N at 240 mm from the LH hanger and the vertical force due to the
down belt is 400N at 720mm from the LH hanger. Find the load on each bearing.
6. A clamp is 120mm long between the centres of the clamping and supporting points.
If the bolt is 45 mm from the clamping point, to what tension must it be tightened up in
order to apply a clamping load of 1000 N? What will be the reaction at the support?
7. A casting is being raised by a 720mm crowbar which is supported at 40mm from the
end where it takes the weight. If the casting has a mass of 612 kg (6000 N weight), and
half this is taken by the crowbar, what force must be applied at the end of the bar in order
to raise the casting?
Blade ,
P " ot l F/xech 40
Pivot
Work'to
he sheared
Fig. 135
8. A bench shearing machine is shown diagrammatically at Fig. 135. If the shearing
strength of steel is 400N/mm 2 , what force F must be applied to shear a piece of material
10mm x 25 mm?
7 Mechanical
principles — II
Friction
Friction plays such an important part in a workshop that the calculations
connected with it are worthy of some consideration. If we press two
surfaces together and attempt to slide one over the other, a resistance is
encountered, and this is caused by the friction between the two surfaces.
Let us consider a block pressed on to a surface with a force W as shown
in Fig. 136. Another force is now applied to the block, tending to slide it
from left to right. As soon as this force is applied, frictional resistance
comes into action and prevents the body from moving. This resistance is
denoted by /and indicated by the arrows at the surfaces. Depending
upon W, and on the nature of the surfaces, however, there is a limiting
value beyond which / cannot increase, so that if we gradually increase
the force tending to slide the block, a point will be reached at which the
block will just be on the point of sliding. Let the force then be F, and since
the block is just about to slide F = /, the frictional resistance. The block
at this point is being acted upon by three forces: (1) W downwards, (2) F
horizontal, (3) the reaction of the other surface (say R). The vector
diagram is shown drawn in Fig. 136 at abc.
F=f Fig. 136
We mentioned earlier, that when friction is neglected the reaction
between two surfaces is normal to their common tangent or along ab in
Fig. 136. It will now be seen that when friction is allowed for, the reaction
moves round so as to oppose motion, and its line makes an angle <j> with
the normal line to the surfaces.
CLAMPING FRICTION 177
This angle ^ is called the Friction Angle and it will be noticed that
. , _ F_ Frictional force
^ ~ W ~ Pressure between surfaces
F
The ratio ^ is called the Coefficient of Friction and is usually denoted
W
by fi.
*=W
F
Thus if Tp = tan <j> and also = /u
tnen + ±
ju = tan <j> .
[The reader will appreciate the fact that the reaction R between the
surfaces moves round to some angular position, and is directed in such a
way as to oppose motion, if he considers where and in which direction it
would act in the event of a definite step being raised in front of the block.]
Approximate Values for the Coefficient of Friction
Nature of surfaces in contact , „ ,T . ^
(Coefficient).
Cast iron on cast iron (dry) 015
020
015
049
056
023
040
033
Ferodo bonded asbestos on steel 03 04
[Values of /u depend to some extent upon the pressure between the surfaces and upon the
speed of sliding.]
Clamping of work
Almost all methods of clamping in the shop depend for their hold upon
the frictional resistance between the two surfaces being clamped. This
applies to work clamped to machine tables, in lathe chucks, work held
in vices, and so on.
Steel on cast iron
(dry)
Steel on brass
(dry)
Cast iron on oak
(dry)
Steel on leather
(dry)
Steel on leather
(greasy)
Oak on oak
(dry)
Leather on oak
(dry)
178
MECHANICAL PRINCIPLES
.15*
Q
100
40
rffe
w:
A\
Work
T¥
t<
10>
w:
7vft
W/////////////7///////////////
Fig. 137
Example 1 . A casting is clamped to a shaping machine table by 4 clamps
arranged as shown in Fig. 137. If the cutting pressure at the tool causes
a thrust of 4500N against the work and if the coefficient of friction
between the work and the table is 015, what is the least tension to which
the bolts be tightened in order that the job may not move under the
pressure of the cut.
If the sliding force to be resisted is 4500N and /u =015, the surfaces
4500N
must be pressed together with a force of 
015
30 000N
Each clamp must therefore exert a pressure of — j — = 7500N
In Fig. 137 the forces exerted by the clamp are shown and T = tension
in bolt, W A and W B = Pressure exerted on work and packing respectively.
We will assume that W A and W B act at the centre of the length being
clamped as shown. Then from W A to the centre of the bolt will be
40  75 = 325 mm and from W B 60mm  5mm = 55mm.
Taking moments about B and working in units of N and mm, we have
Clockwise moment = T X 55
Contraclock moment = W A (55 + 325) = 875 W A .
But W A must be 7500N from above, so that putting in this value and
equating the moments we have
55T = 875 X 7500
T =
875 X 7500
55
 12 000N
Example 2. A milling cutter is tightened up between the collars on the
arbor and is driven by the friction between itself and the collars. If the
FRICTION 1 79
Fig. 138
collars are 20 mm bore and 40 mm outside diameter, calculate the tighten
ing force necessary if the cutter is not to slip when 33 kW is being absorbed
at 70rev/min [Coeff of friction = 015.]
If T newton metres is the torque to tbe transmitted, we have
Power = T newton metres/ second = Ta> Watts
also Power = 33 kW = 3300 W,
Now N = 70 rev/min
cfw 70 x 2tt 22 ,.
so that (o = — Z7 r — = ^ rad/s
OU J
_ Power 3300 x 3 . CAXT
T = = « = 450Nm
oi 22
This torque must be transmitted by the friction between the collars and
the cutter, and slip when it takes place, must occur at two faces:
450
Hence torque transmitted per face = y = 225Nm
This torque will be developed by a tangential frictional force F which we
will assume to act at the mean radius of the collars (Fig. 138).
Hence Mean radius = — = — = 15mm = 00 1 5m
T =Fr,F = = _ 2 A1 5 g N = 15 000N
r 001 5 m
Now if W = Force between the collars
F . 15 000 A1 _
7p= p> » e  w = °' 15
W = l5 ®®° = 100 000N
180 MECHANICAL PRINCIPLES — II
Exercises 7a
1. The tailstock of a lathe has a mass of 21.5kg and the coefficient of friction at the
slides is 0122. What horizontal force will be required to slide the tailstock? Determine
the amount and direction of the least force necessary to slide it.
2. A disc 240 mm diameter has a ring of ferodo 240 mm outside diameter and 20 mm wide
riveted to one face. This is made to press against and drive another steel disc. If the
turning moment to be transmitted is 33 Nm estimate the pressure with which the discs
have to be pressed together. [Take fi for ferodo on steel as 04.]
3. A planing machine is taking a cut on a casting bolted directly to the table. The
force of the cut is 7400N and the job is clamped by four clamps. If the coefficient of
friction between the casting and the machine table is 015, what force must be exerted on
the work by each clamp to prevent the work from sliding under the force of the cut?
4. A 300mm diameter brake drum is attached to a shaft the driving pulley of which is
200mm diameter. When two leatherfaced brake blocks are pressed against opposite sides
of the drum with a force of 200N what force must be applied to the rim of the pulley to
turn the shaft. [Take /u for leather on cast iron as 045.]
5. The table of a planing machine with weight equivalent to 5000N is supporting a
casting which exerts a similar force of 2000N. If "the average speed of the table is 02 m/s
and for half the time a downward cutting force of 800N is acting, calculate the average
power required to overcome friction at the table slides. [Take fi = 008.]
6. A block is clampted between the jaws of a millingmachine vice, the force at the
jaws being 12 000 N. If the cutter operating on this is 80mm diameter and its speed
525 rev/ min, calculate the approximate power being absorbed when the work is caused
to slip in the jaws of the vice by the force of the cut. [Take /u at the vice jaws
as 015.1
Machines and efficiency
A mach ine is a contrivance for receiving energy in some form and con
verting it into energy of a type more suitable for the purpose required.
Most of the energy available in a machine shop is in the form of rotational
energy (line shafting, rotation of driving motors, etc.). A machine, such as
a shaper, receives some of this rotational energy at its pulley, and con
verts part of it into the energy contained in the backwards and forwards
movement of the ram and part into the various other movements required
to transverse the tool across the work.
Each element of a machine may be regarded as a little machine in itself.
For example, in the mechanism for elevating the knee of a milling machine
a torque is applied to the handle, and this torque is converted into the
rotation of a nut or screw. This in its turn raises or lowers the table and
crossslide on the vertical slides. An electric motor is just as much a
machine as any other, for it takes in electrical energy from the mains and
converts it to rotational energy at its driving pulley.
MACHINES AND EFFICIENCY 181
For mechanical machines:
the Effort is the force applied at the input end of the machine,
and the Load is the resistance overcome at the output end.
The ratio p~. is called the Mechanical Advantage of the machine and
+u~ 4 Distance moved by Effort . x , . .„,,,,
the ratio Distance moved by Load m the same time, is called the Velocity
Ratio of the machine.
If we called the effort E, and the load W:
The work input will be E (Distance moved) and the output in the same
time = fF (Distance moved).
The Efficiency of the machine will be = —  —
Input
_ W (Distance moved)
E (Distance moved)
W
But — = Mechanical Advantage
and
E
Distance moved by W 1
Distance moved by E Velocity Ratio
Hence Efficiency = Mechanical Advantage
Velocity Ratio
If the efficiency were unity, the Mechanical Advantage would be the
same as the Velocity Ratio. Actually the efficiency is always less than 1
due to frictional losses, so that the Mechanical Advantage is always less
than the Velocity Ratio.
Example 3. The knee, crossslide and table of a milling machine have
a total mass of 600 kg, and it is found that a force of 60 N must be applied
at the end of the 210mm elevating handle to raise the knee. One turn of
the handle raises the knee 2 mm.
Calculate the Mechanical Advantage, Velocity Ratio and Efficiency
of this mechanism.
Effort (E) = 60N
Load (WO = mg = 600 x 981 = 5886N
Mech Advantage = ,. = 98
60
182 MECHANICAL PRINCIPLES — II
Distance moved by effort
Velocity Ratio =
Distance moved by load
In X 210 mm
2 mm
660
„„, . Mech.Adv. 98 n*Ao 1^00/
Efficlency  Vel. Ratio = 660 = °'' 48 = M ' 8/ »
Example 4. When disengaged from the operating mechanism a grinding
machine table can be pushed along by a force of40N. With the mecha
nism engaged, a torque of 035Nm at the traversing wheel is required to
move it. One turn of the wheel moves the table a distance of 10mm. Find
the efficiency of the traversing mechanism.
The load on the mechanism = resistance of the table = 40N
If 1 turn of the handwheel moves the table 10 mm work done on the
load in 1 turn
= 40N X 10mm = 40N x 0010m
= 04Nm = output
Work done by a torque of 035 Nm in 1 turn
= In X 035 Nm = 22Nm = input
Efficacy <g*£ = <M82 .8.2%
The inclined plane
We have seen in previous work (Fig. 78a) that a screw thread is an inclined
plane wrapped round a cylinder. In order to study the mechanics of the
screw, therefore, we must give some attention to the inclined plane. We
find that in doing this friction plays an important part and its effects must
be allowed for.
Tightening up
When a nut is being tightened up under a load W, the conditions are equi
valent to pushing a weight W up an inclined plane sloping at the helix
angle of the screw. The force F which is pushing W is the tangential force
at the mean radius of the screw and is being applied in a horizontal direc
tion. This is shown in Figs. 139 and 140.
Now if there were no friction between the block and the plane the
SCREW THREAD AS AN INCLINED PLANE 183
Load on
Nut(W)
Effort
(F)
Angle of Incline*
Helix Angle of Screw
Development of Screw and Nut
Fig. 139
vector diagram for the forces acting on the block would be as abc; ab =
weight of block acting downwards, be = F, the force to pushing it up the
plane and ca = the reaction between the plane and the block. We have
seen, however, that when friction is present, the reaction R is no longer
perpendicular to the surfaces, but is rotated round through the friction
angle <j> in the direction to oppose motion. Hence the vector for R will
be rotated round to ad, and the force F required to push the block up the
plane will become = bd.
Lead of
Thread
j? c
Fig. 140
bd /\
Now — = tan bad = tan (or + <j>)
ab
W
tan(ar + ^) and F = Wtan(a + ^)
(1)
184
MECHANICAL PRINCIPLES — II
In pushing the block from A to B, F moves the distance AC and W
is raised to distance BC.
Hence the input = work done by F = F.AC
and the output = work done on W — W.BC
The efficiency {rj)
output
fF.BC W 4 I BC
r . ^ = 77 tan a since
F.AC F
tan or
input F.AC F L" AC
But from above: F = W tan (a + (f) and if we substitute this for F we
have
Wtana
tan rv —
w.
t] = p tan a =
Wtan(a + 0)
tana
a = Helix angle of screw, i.e. tana =
<j> = Friction Angle
tan (a + (j>)
Lead
(2)
Mean circum
Tan = /u
This is an expression for the efficiency of a screw when tightening up and
is in terms of /u and the helix angle of the screw.
Loosening
When a nut is being unscrewed it is equivalent to the load Amoving down
the plane. There are two cases to consider: (1) When is greater than a
and the load must be pushed down, and (2) when a is greater than </> and
a push up the plane is necessary to stop W from sliding down of its own
accord.
(1) greater than a (Fig. 141).
As motion is now taking place in the opposite direction, the reaction
line ad swings round to the other side of the normal ac, and the vector
d ]? b c
Fig. 141
FRICTION AT VEE THREADS
bF d
Fig. 142
diagram is abd. In this case since <j> is greater than a, ad swings lfound to
a position to the left oiab.
Force required at mean radius to unscrew nut
= F = db — ab tan had
= W tan (</>  a) (3)
(2) a greater than ^ (Fig. 142).
In this case a force = bd will be required to hold the nut fjom un
screwing.
F = bd = ab tan(a — f)
= Wtan(a  </>)
185
(4)
Vee threads
The above reasoning for screw threads assumes that W is the normal load
on the thread surfaces. Whilst this is true for a square thread [Fig. 143 (a)] ,
when a vertical load W is applied to a vee thread the force is resolved into
components as shown by the vector diagram abc shown at (b).
(b) Fig. 143
186 MECHANICAL PRINCIPLES — II
The normal load on the thread surface is now the length ac, whilst be is
the force tending to burst the nut.
W
Hence normal load on thread face — ac — = W sec 6 .
cos 6
As the frictional resistance depends on the normal pressure we may
modify our previous formulae to apply it to vee threads by modifying the
value of the coefficient of friction (/u).
If, instead of taking /u we take for it a value equal to /u sec0, we may use
the previous formulae as they stand.
For a metric thread, the veeangle is 60°, hence 6 = 30° and sec# =
1155.
Hence for a metric thread take a modified value for the coefficient of
friction equal to 1155 ,u.
Friction at a nut face
In addition to the friction at the threads, friction at the nut face must be
allowed for.
Let W = load on nut
Hi = coefficient of friction
r = mean radius of nut.
Then frictional force at nut face = fxW.
Torque necessary to overcome this = (force) (radius) = fiWr
Example 5. A screwoperated arbor press has a squarethreaded screw
50mm dia 5mm pitch single start. If the coefficient of friction at
the threads is 0 10, what load may be applied by the press when an effort
of 200N is applied at the end of a handle 200 mm long attached to the
screw.
Since tan ^ = /u; tan <j> = 01 and the friction angle (0) = 5°43'
The mean circumference of the thread = ,t(50  25) = 475 n =
1492 mm
If a is the helix angle of the thread
Lead 5 nm»c
tana = ~ = . An . = 00335
Circum 1492
from which a = 1°55'
The mean radius of the screw = ^ = 2375 mm so that a force of
FRICTION AT A NUT FACE 187
200N at 200mm radius will be equivalent to a force — — — at 2375
mm radius = 1684N. This is the force F up the inclined plane.
Now F = Wtan(« + ^)(Fig. 140)
= fTtan(l°55' + 5°43')
= W tan7°38' = W X 01340
w  orao = 515? = i^ON
Example 6. A machine slide weighing 255 kgf (2500 N) is elevated by a
2start acme thread (29° thread angle) 40mm dia, 4mm pitch. If the
coefficient of friction is 012, calculate the torque necessary (a) to raise the
slide, {b) to lower it. The end of the screw is carried on a thrust collar,
32 mm inside and 56mm outside diameter.
Mean dia of thread = 40 mm  2 mm = 38 mm
Mean circum = 38 n = 1194 mm
Lead = 2 x 4 mm = 8 mm
g
If a = helix angle tan a = rnrz = 00670
a = 3° 50'
Since the thread half angle = 14^° we have to use a modified coefficent
of friction = 01 2 sec 14}°
= 012 x 1033 = 0124
tan^ = 0124 and ^ = 7° 4'
To raise the load we have if F = force at the mean radius
F = Wt&n(a + f) = 2500 tan(3° 50' + 7° 4')
 2500 tan 10°54' = 2500 x 01926 = 4815N
The torque will be F (radius at which it acts)
= F x 38mm = f x 0019m = 4815N x 0019m = 915Nm
2
To this torque must be added the friction torque at the thrust collar.
Frictional force = 2500^ = 2500 x 012 = 300N
Frictional torque = 300 (mean rad of collar)
= 300 x ^y5B. = 300N x 0022 m = 66Nm
Total torque to raise slide = 91 5Nm + 66 Nm = 1575Nm
188 MECHANICAL PRINCIPLES — II
To lower the load F = JFtan(0  a) (Fig. 141)
= 2500 tan(7°4'  3° 50')
= 2500tan3°14' = 2500 x 00565 = 14125N
Torque = 14125N x 0019m = 268Nm
Adding the friction torque at the collar we have
Total Torque to lower the slide = 268 Nm + 66 Nm = 948Nm.
Friction of sliding keys
It is often necessary in machinetool construction to slide a collar or wheel
along a shaft at the same time as the wheel is being driven by the shaft
through one or two sliding keys. The force required to slide the collar or
wheel along the shaft is worthy of "consideration as it is affected by the
disposition of the keys.
In Fig. 144(a) the key is fixed in the shaft which rotates in the direction
of the arrow and drives the outer part. Clearances have been exaggerated
and it will be seen that the torque is transmitted by a force F at the key
and a similar force W acting at the circumference of the shaft. If r is the
radius to the centre of the key as shown, and T is the torque being trans
mitted, then T = Fr and F = ~
r
The force required to slide the outer member along the shaft will be
F/jl + Wfi= IFfx (since F = W).
Fig. 144
FRICTION OF SLIDING KEYS 189
In Fig. 144 (b) there are two keys, and if they are well fitted so that
each takes an equal share of the load, the torque is given by T = F x r +
W x r= F,{2r) [since F l = W,].
T
Hence F, = ^, which is half the force for the case of Fig. 144 (a).
Force to slide outer member along the shaft = F y /u + W r /u = 2F 1( u,
which is half of that for the case of Fig. 144 (a).
Exercises 7b
1. In a hand crane a 20T gear attached to the handle drives an 85T gear on the rope
drum. The radius of the drum to the centre of the rope is 35 mm. If the handle is 300mm
long, calculate the velocity ratio. If the efficiency is 75%, what force must be applied to
the handle to raise a load of 250 kg on the rope?
2. A chain conveyor carries goods up an incline of 28° at a speed of 02m/s. If the average
mass of the articles carried is 20 kg, and they are spaced at 300 mm centres, calculate the
power necessary to drive the loaded conveyor if its efficiency is 75%, the incline carries
60 articles, and 04 kW is necessary to overcome friction.
3. The saddle of a lathe is equivalent to a weight of 1600N and 1 turn of the traversing
wheel moves it 100mm along the bed. If the efficiency of the traversing gear is 07 and
the coefficient of friction at the slides 010, calculate the force necessary at the rim
of a 140mm wheel to move the saddle.
4. A flypress has a screw of 50mm lead, the efficiency of which is 60% . Neglecting the
weight of the screw and top arm, what force must be applied at the end of and per
pendicular to the level of 280mm radius to put a force of 2500N on the ram?
5. Calculate the efficiency of a 20 mm squarethreaded screw of 5 mm pitch 2 start if
the coefficient of friction at the threads is 0080. What tension may be exerted on this screw
by a nut if 120N is applied at the end of a 350 mm spanner? [Neglect friction at the nut
face.]
6. The efficiency of a screw and nut is 15%, and the coefficient of friction at the nut
face is 01. If the lead of the thread is 2mm what force must be applied to the end of
a 240 mm wrench to pull up the nut against a tension of 8000 N? [Mean radius of nut =
15mm.]
7. The spindle of a lathe is connected to the leadscrew by the following gears: =r: =
30 20
"sn x Z?" ^ l ^ e l ea d screw i s 5 mm pitch, calculate the velocity ratio between the carriage
and a point on the rim of an 160 mm chuck screwed on the spindle. If the overall efficiency
of the arrangement is 10%, calculate the force necessary at the rim of the chuck to turn
the lathe and traverse the carriage against a resistance of 200N.
8. Calculate the efficiency of a M24 thread when the coefficient of friction at the
threads is 008. [For the M24 thread take the mean diameter as 22 mm and the pitch 3 mm.]
9. Calculate the work done in pushing a slide of mass 100 kg up an inclined plane
sloping at 30°, if the coefficient of friction is 015, the plane is 3 m long, and the push is
applied horizontally. What horizontal effort would be necessary to hold the slide from
moving down the plane?
190
MECHANICAL PRINCIPLES
10. A wheel slides along a shaft and is driven by a sliding key. If the shaft is 45mm
diameter and the key projects 5 mm from the circumference, calculate the force necessary
to slide the wheel along the shaft when 66 kW is being transmitted at 315rev/min and
H = 015.
11. If the shaft in the last example were fitted with two opposite keys projecting the
same amount as before, calculate the torque being transmitted when an axial force of 60 N
is required to slide the outer member along the shaft, i/n = 01 5.]
Bearings
The main function of a bearing is to hold and line up the shaft it carries
and support the load to which the shaft is subjected. Bearings are generally
designed on the basis of the load carried per unit of projected area, and
if the length and diameter of a bearing are / and d respectively, the
projected area will be / x d (Fig. 145).
Fig. 145
If the load on the shaft = W, the intensity of bearing pressure (p) will be
Load
Area
W
P= ld
The pressure to which bearings may be subjected in practice depends
upon various factors, including the speed, method of lubrication, dura
tion of full load operation, materials in contact, and so on.
191
The following table conveys an idea of bearing pressures used:
Table of Bearing Pressures
Allowable
Type of Bearing
Pressure
N/mm 2
Lineshafting (bronze lined)
07 10
Highspeed engines: Main bearings
10 20
Crank pins
20 40
Gas engines: Main bearings
35 50
Crank pins
10 125
Punching and shearing machines (low
speed intermittent
loading)
15 30
Horizontal turbines
03 05
Probably the most severely loaded bearing in the whole of engineering
practice is the tailstock centre of a lathe.
Let us consider the vertical tool pressure only, and assume the moderate
case of a load of 4000 N, with the centre in a hole measuring 5 mm at
the large diameter of the countersink. Since the angle of the countersink
is 60° the projected area will be
i X 5 x 0866 x 5 = 108 mm 2
and the bearing pressure when the tool is cutting close to this centre will be
4000N
r^r r = 370 N/mm 2 , about half the ultimate failing stress of a good
108 mm 2
quality mild steel ! The main bearing carries only the same load with an
area of probably 3000 to 4000 mm 2 .
Example 7. If the bearings for the shaft in Example 12, p. 171 are to
be proportioned so that their length shall be 2\ times the diameter,
calculate their dimensions if the bearing pressure is not to exceed
04 N/mm 2 .
The biggest load to be carried = 460 N
If L = 2\d, L x d = 2W x d = 2\d 2 = Area of bearing.
^ <• . • Load 460 nm ,
But area of bearing = =, = 77r = llSOmnr
b Pressure 04
Hence 2\d 2 =1150
<P = il^ = 460mm 2
192 MECHANICAL PRINCIPLES
d = \/460 = 215 mm
L = 2\ x 215 = 538 mm
Bearing friction
When a bearing is properly lubricated the two metals forming it are not
in contact but are separated by a thin film of oil. The friction now is
not that of one metal rubbing on another but is the internal friction of the
lubricant itself. It has been found experimentally that the coefficient of
friction in a bearing with film lubrication depends upon the rubbing
speed and upon the pressure, and the relation between them is
P =
KV\
where /u = coeff. of friction
v = surface speed of shaft in m/s
p = bearing pressure in N/mm 2 (i.e. MN/m 2 )
K = a Constant = 0032 for the usual oils.
Work lost in bearing friction
The effect of friction in a bearing is to introduce a tangential resistance
at the periphery of the shaft. If W = load on the bearing and n = co
efficient of friction, then the tangential resistance will be W = /u. Calling
this resistance F, we have F = W/u (Fig. 146).
The work lost per second will be Fv
and the power lost = Fv watts
Fig. 146
STRESS AND STRAIN 193
Example 8. A 50mm dia shaft running at 525rev/min carries a load of
8000N.The bearing is 100 mm long. Estimate (a) the coefficient of friction,
(b) the tangential friction resistance, and (c) the power lost in friction.
w u *u ♦ 0032Vv
We have that a =
P
v = J^L = 22 >< 50 >< 525 = 1.375m/s
60 000 7 x 60 000
8000
Bearing pressure (p) = ^ r^r = 16 N/mm 2
0032 x W375 0032 x 1172 _ 00234
^~ 16 ~ 16
Tangential frictional
resistance = 00234 x 8000
= 187 N
Power lost in
friction = 187 N x 1378 m/s
= 257W
Stress and strain
A material is placed in a state of stress when a load acts upon it, and the
numerical value of the stress is given by
Load acting F , FT j a a a
r: . p , — r = r, when F = Load, and A = Area
Area subjected to load A
The SI unit of stress is therefore one unit of load divided by one unit of
area, i.e. the newton per square metre (N/m 2 ). This unit is very small
for practical purposes and so stresses will often be quoted in meganewtons
per square metre (MN/m 2 ). This unit is quite convenient, since
lMN/m 2 = lN/mm 2
and as loads are often quoted in newtons, while dimensions of engineering
components are usually quoted in millimetres, if we ever require a stress
in MN/m 2 it is useful to evaluate the stress in N/mm 2 , e.g.
45 N/mm 2 = 45 MN/m 2 .
194 MECHANICAL PRINCIPLES — II
It is possible that the unit adopted for fluid pressure will be the bar, a.
unit inherited from the previous metric system. In some ways this is a
more convenient unit since it utilises the cm 2 as the unit of area.
1 bar (b) = da N/cm 2 = 10 N/cm 2
1 hectobar (hb) = 10 3 N/cm 2
As a mental landmark, the bar is very near to atmospheric pressure:
1 bar = 10 N/cm 2 1 atm = 1013 N/cm 2
The weight of a body is the effect of gravity on its mass. The effect of
gravity on a mass of m kilogrammes can be taken as,
weight = mass x acceleration due to gravity
= mass x 981 newtons.
As we discussed in Chapter 1 , the reader will find, in his normal life,
that the mass of 1 kilogramme is referred to as "weight" or a kilogramme
of mass and written 1 kg, with the inference of weight.
In our work it is sometimes expedient to express the weight of 1 kg of
mass in its kg form and when doing this it should be written kgf . At the
same time it should be remembered that 1 kgf =981 newtons.
When materials are stressed they change their shape: for example, a
bar will lengthen under tension or shorten under compression. This
change of shape is called strain. We usually express the strain in terms
of the natural length ofthe material, so that if abar 100 mm long is subjected
to tension and stretches 01 mm, the strain is expressed as j^ = 0001.
Within certain limits the materials with which we have to deal behave in
an elastic manner, i.e. the deformation caused by a load vanishes
when the load is removed. If, however, the load on a bar is gradually
increased, a point is reached beyond which the material will not return
to its original shape when the load is removed. This point is called
the Elastic Limit ofthe material. For most materials it has been found that
within the elastic limit the change in length is proportional to the load
producing it: e.g. if 1000 N causes an elongation of 005 mm, 2000 will
cause 010 mm, and so on. Hence we may say that within the elastic
limit:
Stretch is proportional to Load
STRESS AND STRAIN 195
or, since for the same bar, strain is proportional to stretch and stress to
load:
Strain is proportional to Stress.
Stress
This is the same thing as saying: ~ — — = a constant quantity
This constant quantity is called Young's Modulus, and has a particular
value for every material. It is usually denoted by the letter E. For
steel E has a value of about 200 000N/mm 2 .
It is difficult at first to visualize and to appreciate what E = 200 000
N/mm 2 signifies. The following way of considering it might help the
reader:
Stress
t= — — means Stress per unit Strain and a material would have unit
Strain
strain if its length were doubled.
(Original length = 1 ; Stretch = 1 ; Strain = f)
If, then, a material could remain elastic whilst its length were doubled
under a load, the stress in the material would have the value E.
Example 9. A 20 mm bolt 160 mm long carries a load of 20 kN. Calculate
the extension in the bolt if E = 200 000 N/mm 2 .
• *U U H L ° ad 20 000 20000 « 1 XT/ 2
Stress in the bolt = r = = , . = 637 N/mm 2
Area ti ,_ rw? ji^'Z
4"(20) 2
Extension Ext
Strain =
\ 200000 =
Extension =
Orig. length 160
„ Stress
E = =. — —
Strain
E = 200 000 Stress 637
637 637 x 160
Ext Ext
T60
637 x 160 _ 005 lmm
200 000
Example 10. A 20 mm steel bolt is threaded through a brass sleeve 100 mm
long, 24mm bore and 32mm outside diameter. A nut and washer are
196 MECHANICAL PRINCIPLES — II
put on and the nut tightened up until the brass sleeve has shortened by
005 mm. Calculate the extension in the bolt.
E for steel = 200 000 N/mm 2 E for brass = 80 000 N/mm 2
Strain in sleeve = ^ = 00005
and since E ===^: Stress = ^(Strain) = 80 000 X 00005 = 40N/mm 2
Strain
Stress in sleeve = 40 N/mm 2
Crosssectional area of sleeve = (32 2  24 2 )  ==(1024  576)
22 x 448 a „ 2
= ^ = 352 mm 2
Hence compressive load carried by sleeve = (Stress) (Area)
= 40 x 352 = 14 080N
This will be equal to the tension in the bolt.
•11* 14080 14080 AA OXT/ 2
Hence stress in bolt = = ,. . ^ = 44 8 N/mm 2
p*y 3142
„ Stress , Ci . Stress 448
E = = — — and Strain =
But strain =
Strain E 200 000
Extension Ext
Orig Length ~ 100
 1^L_ 448
' • 100 ~ 200 000
100 x 448 nM ~ A
Ext = 200 000 = ° 0224mm
Extension of bolt = 00224 mm
Exercises 7c
1. A bearing has to carry a load of 3000N, with a bearing pressure of 075N/mm 2 .
I f the length of the bearing is to be made equal to twice its diameter determine its dimensions .
2. A 50mm diameter shaft runs in two bearings spaced at 25 m centres, the bearings
each being 80mm long. Loads of 700N, 800N and 750N act on the shaft at 05m, l25m
and 1 8 m respectively, from the LH bearing . Determine the load and bearing pressure at each
bearing.
3. A bearing 50mm diameter, 80mm long, carries a total load of 6000N. If the shaft is
rotating at 210 rev/min estimate the coefficient of friction from the expression /u —
0032Vv
. Hence determine the number of joules of work lost in friction per second,
P
i.e. the power lost in watts.
STRESS AND STRAIN 197
4. The end thrust on a spindle is taken by a collar on a 60mm diameter portion of the
shaft. If the maximum thrust is 2100N, and the bearing pressure is not to exceed 07 N/mm 2
determine the necessary top diameter of the collar.
5. A line of 60mm shafting runs in six bearings each 100mm long. The average load
on each bearing is 6000N and the speed of the shaft is 3 1 5 rev/min . Estimate the coefficient
of friction and the power being lost in friction.
6. If the Elastic Limit of a certain material were at 160 N/mm 2 , what load would a
20 mm diameter bar of the material carry without sustaining a permanent stretch?
7. Taking the ultimate stress of a mild steel to be 500N/mm 2 , calculate the load in
Newtons necessary to fracture an M6 metric bolt at the root of the thread. [Take the
diameter at the root of the thread as 45 mm .] Hence taking the efficiency of the
thread at 10%, find what force at the end of a 200mm spanner will cause the bolt to
fracture. [M6 thread has a pitch of 1 mm]
8. A drawbolt 20mm dia, 800mm long, is pulled up to a tension of 16 000N.
Calculate the stress in the bolt, and if E = 200 000 N/mm 2 , determine the total
extension.
9. An air cylinder is 140 mm diameter and the cylinder head is held on by six M 12 studs
and nuts. If the nuts are tightened up to an initial tension of 1000N, calculate the stress
at the root of the thread when the air pressure in the cylinder is 06N/mm 2 . [Take the root
diameter of the M12 thread at 95 mm.]
10. A steel ring 20mm wide, 10mm thick and 14985 mm inside diameter, is heated up
and shrunk on to a shaft 150mm diameter. If E = 200000N/mm 2 , estimate the stress and
the tension in the ring.
8 Mechanical
principles  III
The equations of motion
The distance travelled during a given time by a body moving at a certain
constant speed will be given by multiplying the speed by the time, or if
s = space travelled, v = velocity (or speed), and t = time,
s = vt (1)
It is important that the time units of v and t are coherent (e.g. if v is in
metres/ second, then t must be in seconds and s will then be in metres).
Acceleration
Acceleration is the rate of increase of velocity. For example: if a body
starts from rest with an acceleration of 1 m/s every second (i.e. 1 m/s 2 ),
its velocity at the end of 1 second will be 1 m/s, at the end of 2 seconds it
will be 2 m/s, and at the end of / seconds it will be t metres per second.
Hence we may say for a body starting from rest with an acceleration a:
Final velocity (v) after time t ; v = at
If, instead of starting from rest, the body already had an initial velocity
of u, then: Final velocity after time t = u + at
This gives us v = u + at (2)
Instead of accelerating, a body may be slowing down or decelerating.
Its acceleration will then be a minus quantity, and we shall have for its final
velocity:
v = u — at
This slowing down is generally termed retardation.
A graph of velocitytime for equation (1) above is shown in Fig. 147(a)
and the area under the graph is equivalent to the distance moved.
If we now plot a similar graph for equation (2) it will be as Fig. 147(6).
In this case, due to the acceleration, the velocity is increasing at a constant
rate and the graph is a sloping line instead of a horizontal one. The
distance moved will be, as before, the area under the graph (Area O ABC).
ACCELERATION 199
Fig. 147
But OABC = OADC + ADB
s = ut + \tat
This gives us s = ut + hat 2
If u = O then the line starts at O and
s = hat 2
(3)
The reader is strongly advised to interpret problems of accelerated
motion as far as possible with the help of a graph.
Equation (3) could have been obtained as follows: For any motion the
space travelled = (average velocity) (time). In Fig. 147 (b) the average
velocity is at a height midway between A and B, i.e. u + \at. Multiplying
this by the time t we get s = ut + hat 1 as before.
Finally, we require an equation connecting v, s and a, so we must
eliminate t.
and
In equation (2) we have v = u + at
v — u
t =
a
Substituting this value for t in equation (3) we have
s = ut + hat 2
uv  u 2 , /v 2  2uv + u 2 \
^r + *\ — ? — ;
200 MECHANICAL PRINCIPLES — III
Putting on the common denominator 2a
2uv — 2u 2 + v 2 — 2uv + u 1
s =
s
2a
u 2
2a
which gives us that v 2 — u 2 = 2as (4)
If the initial velocity u = (body starting from rest)
then v 2 = las.
In problems dealing with the motion of bodies falling under the action
of gravity then,
a = acceleration due to gravity = 981 m/s 2
This is generally signified by g instead of a, and the reader should note
particularly that in using 981 forg the units are in metres and seconds. The
units of all the other quantities in the equations must be kept in coherent
units.
Example 1 . A drop stamp falls freely for 6 m under the action of gravity.
Find its velocity at the moment it strikes the tup.
Here u = s = 6ma = 981 m/s 2
and v 2 = las
= 2 x 981 x 6 = 11772
v = Vl 1772 = 1085m/s
Example 2. A shaping machine ram is running on a stroke of 450mm. It
starts from rest, accelerates at a uniform rate until the centre of the stroke
and then retards at a unifrom rate to a standstill at the end of the stroke. If
the stroke occupies \\ second, find the acceleration and the maximum
speed attained.
This problem is best illustrated graphically, and the motion is repre
sented in Fig. 148.
The ram is accelerated from O to A, its velocity increasing uniformly
and the reverse process takes place from A to B.
The area OAB represents the space travelled, which in this case is
450 mm = 045 m
ACCELERATED MOTION 201
Hence
045 = £OB.AC and since OB = 1£ seconds
045 = i.HAC = AC
.. AC = v = 06 m/s
For half the stroke v = at
06 = a. \ anda = 08m/s 2
Hence acceleration = 08 m/s 2
max speed of ram = 06 m/s
Note that the average speed of the ram = 03 m/s
Vel.
Fig. 148
Fig. 149
Example 3. A planing machine table is set for a travel of 900 mm. It starts
from rest, accelerates uniformly during the first 1 50 mm, runs at a constant
speed for 600 mm and then retards uniformly to rest during the last 1 50 mm
of the travel. The total time taken travel the 900mm is 4 seconds.
Find (a) the average speed, (b) the maximum speed, (c) the acceleration.
A graph showing the motion is shown at Fig. 149. Let the maximum
velocity be v and the times of acceleration, uniform speed and retardation
f,, t 2 and tj, as shown.
(a) The average speed will be
distance 09 m
= 0225 m/s
time ~~ 4s
Since the first and last portions of the travel are identical: retardation =
acceleration (a).
For the first part of the stroke:
v 2 = las = 03a (since s = 015 m) (1)
202 MECHANICAL PRINCIPLES — III
Also, for the first and last portions of the travel
v = at, = at, and t, = t, = —
a
and for the second s 2 = vt 2
t 2 — — (since s 2 = 600 mm = 06 m)
.,,,,, v , 06 v . 2v 06 ,«v
..t l +t 2 + t i = + — +  = 4 = — + — (2)
a v a a v
(since total time = 4 seconds)
v 2
From equation (1): # = —
Substituting this in equation (2):
2v x 03 06 _
V 2 V
06 06 . .
— + — =4 i.
V V
and
4v= 12, v
Since from (1)
v 2 03 x 03
a ~ 03 ~ 03
ii=4
v
= 03m/s 2
Exercises 8a
1. A shaft starts from rest and with uniform acceleration attains a speed of 300 rev/min
in half a minute. Sketch the graph of velocitytime, and calculate (a) the acceleration in
rev/min 2 ; (b) the number of revolutions made by the shaft during the period.
2. A drop stamp falls freely under the action of gravity from a height of 8 m. Calculate
(a) the time of fall, and (b) the velocity at the instant it strikes the bottom block.
3. Starting from rest, a shaping machine ram, with uniform acceleration, reaches a
speed of 24 m/min during 240mm of travel. Find the acceleration and the time taken.
4. A cam rotates at 180 rev/min. During 90° of its revolution it causes a plunger to
rise a distance of 25 mm. Half the rise is made with uniform acceleration and the remainder
with an equal retardation. Calculate the acceleration of the plunger and sketch the graph of
velocitytime for it.
5. The total travel of a planer table is 21 m. Starting from rest, the table accelerates
uniformly for \\ seconds, runs at a constant speed for 2 seconds, and retards to rest
during 1 seconds. Calculate the acceleration, and the speed during the middle interval.
Sketch the graph of velocitytime.
6. After the power has been shut off, a flywheel rotating at 180 rev/min slows down
to rest during 90 turns. If the retardation is uniform finds its value in rev/min 2 and the
time taken by the wheel in coming to rest.
7. A machine slide has an acceleration of 008 m/s 2 . How far will this slide travel from
MASS AND WEIGHT 203
rest before reaching a speed of 24m/min, and how long will it take for this to be effected?
8. A slide starts from rest and travels a distance of l25m with an acceleration of
01 m/s 2 , and then comes to rest again with a retardation of 005 m/s 2 . Calculate (a) the
maximum speed attained, {b) the total time taken.
Sketch the graph of velocitytime.
Motion and force
Newton's First Law of Motion states that every object remains at rest or
moves with uniform velocity in a straight line until compelled by some
force to act otherwise. His Second Law states that change of motion is
proportional to the impressed force, and takes place in the direction in
which that force acts.
Before we can arrive at a quantitative relation between force and
change of velocity we must consider the velocity of an object as something
more than a rate of movement, since force required will depend upon the
size or amount of material in the body.
Mass and weight
Every object is made up of a mass of material (iron, wood, stone or what
ever it might be), and this mass is constant and invariable, so long as we
do not cut away from or add material to it. Due to the gravitational
pull of the earth acting on this material every body exerts a downward
force. This force is the weight of the body. The weight of a body is not a
constant quantity because the pull of gravity varies according to the
distance from the centre of the earth. For example, the weight of the same
amount of mass would be about \ percent greater at the poles than at the
equator. We thus have the mass of a body, being the amount of substance
in it, and a constant quantity, and the weight, being the downward force
caused by the gravitational pull acting on the mass.
We may now define the quantity of motion of a body as being (mass)
(velocity), and for the given velocity this will not vary since mass is a
constant quantity. This product mv (m = mass) is called Momentum.
By Newton's Second Law: Force is proportional to change of motion,
i.e. Force is proportional to change of mv.
But m cannot change as it is constant.
.. Force varies as m (change of v)
Now change of velocity is acceleration.
Hence Force varies as ma .
If the units of F, m and a are chosen so that unit force causes unit
204 MECHANICAL PRINCIPLES — III
acceleration on unit mass we may say F = ma, and this is the fundamental
relationship between the quantities.
The SI system of units is coherent. This means that the SI unit of force
is the force that gives to one unit of mass one unit of acceleration. To
honour the contribution of Sir Isaac Newton to Science this is called the
newton. Hence
F (newtons) = m (kilogrammes) X a (metres per second every second)
Let us now consider the action of gravity on a mass of m kilogrammes,
i.e. the weight of the body.
Using F = ma
a = g
and F — mg
The weight of a body of mass m kilogrammes is, therefore, mg newtons;
g may be taken as 981 m/s 2 .
Example 4. If the planing machine table in Example 3 has a mass of
400 kg, and the coefficient of friction at the slides is 01, calculate (a) the
total force necessary to accelerate it, and (b) the power being taken to
accelerate and overcome friction at the instant when the table has moved
75 mm from the beginning of its travel.
(a) Force to overcome friction = (W)/u
= (mg)fi = 400 x 981 x 01 = 3924N
Force to accelerate = ma
= 400 X 03 = 120N
Total force = force to overcome friction + force to accelerate
= 3924 + 120N = 5124N
(b) After 75 mm of movement, speed of table = 1 = 03 m/s
Rate of doing work = (force) (speed)
= 5124 x 03 = 15375W
Power = 154W = 0154kW
We might add that during the middle portion of the stroke no acceleration
is taking place and the table has merely to be kept moving against friction
(neglecting any cutting force).
Here we have: frictional resistance = 3924 N
FORCE OF HAMMER BLOWS 205
Speed = 06 m/s
Power = 3924 x 06 = 23544 W = 0235 kW
Force of hammer blows
The force of a hammer blow may be estimated by considering the retarda
tion of the hammer as shown by the following example:
Example 5. A hammer of mass 1kg and moving at 2 m/s strikes a pin
and is brought to rest by driving in the pin 5 mm. Calculate the average
force of the hammer blow on the pin.
To find the retardation of the hammer we may use the equation
v 2 — u 2 = las
Here v (final velocity =
u (initial velocity) = 2 m/s
s = 5 mm = 0005 m
— u 2 = las
I 2
a = —
1 x 0005
Force = ma = 1 X 400 = 400 N
= 400 m/s 2
Example 6. A machine slide, of mass 500 kg and moving at 24m/min, takes
1 second to come to rest after the power is shut off. Calculate the average
Motional resistance assumed as a force acting against the slide.
To find the retardation of the slide we may use the formula
v = u + at
Here v = 0: « = £= 04 m/s: t = 1 second
= 04 + a(l) a = 04 m/s 2 (ve because retardation)
If the average resistance is denoted by F
F = ma = 500 x 04 = 200N
Exercises 8b
1. A slide of mass 100kg starts from rest and accelerates uniformly to a speed of
12m/min in 2 seconds. Neglecting friction, calculate the force necessary to produce the
acceleration.
2. A machine table has a mass of 60kg and the frictional resistance at the slides is
206 MECHANICAL PRINCIPLES — III
equivalent to a force of 50N. If a constant force of 80N is applied to this slide, what
speed will it have attained after 2 seconds?
3. If the planer table in Ex. 8a, No. 5, had a mass of 300kg, and the coefficient of
friction at the slides were 008, what force would have been required to accelerate it
during the first portion of its travel?
4. A lkg hammer head moving at l2m/s is brought to rest by driving a pin through
a distance of 10 mm. Assuming the retardation of the hammer to be constant, calculate
the average force of the blow delivered to the pin.
5. Taking the mass of the cam plunger in Ex. 8a, No. 4, to be 12 kg, calculate the
vertical force exerted by the plunger on the cam face during the portion of the lift that
the plunger is accelerating. [The plunger moves upward on a vertical centre line.]
6. A shaping machine ram has a mass of 150kg and on its return stroke it starts from
rest and moves through a distance of 360 mm in 075 s with uniform acceleration. Neglecting
friction, calculate the acceleration, the accelerating force, the velocity after the ram has
moved 360mm and the power necessary to move it at that instant.
7. A drop stamp of mass 100kg falls freely from a height of 6m. Calculate its final
speed. It is brought to rest by compressing the metal on the bottom block through a
distance of 10mm. Determine the retardation of the stamp and the average force ofthe
blow delivered .
8. A machine slide of mass 50 kg is moving at 08m/s when the power is shut off. If the
frictional resistance to its motion is equivalent to a force of 50N, how far will it travel
before coming to rest?
Energy
There are various forms of energy (e.g. heat energy, chemical energy,
electrical energy, etc) and the origin of all of them may be traced back
to energy derived from the heat ofthe sun. The Law ofthe Conservation
of Energy states that energy cannot be destroyed: one form may be
changed into another, but we can neither create new energy nor destroy
that which is in the universe. At the moment we are interested in the
energy of mechanical movements, and in this connexion we may define
the energy of a body as the power of overcoming resistance or of doing work.
A body may possess energy by virtue of its position, e.g. a weight on
a cord wound round a shaft may rotate the shaft and do work as it descends
to the floor. A woundup clockspring possesses energy. Energy of this
kind is called Strain Energy.
When a body is moving, it possesses energy of a different kind called
Kinetic Energy. A rotating flywheel or an oscillating machine slide are
examples of moving bodies possessing kinetic energy.
Since we may convert energy from one form to another without loss,
we may obtain an expression for kinetic energy:
Let a body of mass m (i.e. weight mg) be raised to a height h above
some datum line (Fig. 150). When in this position the body will possess
ENERGY
207
mgh units of potential energy, since this amount of work must be expended
to lift it there. If now the body be allowed to fall, it will lose its potential
energy, and gain an equal amount of kinetic energy and
KE = PE = mgh
We may express the K.E. in terms of v because
v 2 — u 2 = las.
u = 0; a = g, and s — h.
:.v 2 = 2gh
and
Hence KE = mgh
2g
v 2 mv 2
mgx Tg = ~T
(6)
An expression for kinetic energy in terms of m and v.
The SI unit of all forms of energy is the newton metre, i.e. the joule (J).
The reader is reminded that energy and work are interchangeable: to
put a body in possession of a certain amount of energy requires the
expenditure of an identical amount of work.
m
1 1
1 F = \mg
v\\ I
1 1 1
Datum L——L.UPS—1
Fig. 150
Example 7. A shaping machine ram has a mass of 200 kg and accelerates
uniformly from rest, covering the first 360 mm of its travel in 075 s. Find
the velocity at the end of this travel, and show that the kinetic energy of
the ram at that point is equal to the work done in accelerating it.
In Fig. 151 the motion of the ram may be represented by the lineOA,
and AB represents the velocity v attained after 075 second.
208 MECHANICAL PRINCIPLES — III
Since space travelled = Area OAB = 360 mm = 036 m
i(v X !) = 036
and 5" =036, v = 096 m/s
o
v
To find the acceleration we have v = at or a = 
096 ne / 2
0T75 = 1  28m/S
Force = ma = 200 x 128 = 256 N
Work done = (force) (distance) = 256 x 036 = 92 16 J
Energy of 200kg ram at 096 m/s = ^ = 200 x °' 962 = 9216 J, as before,
Example 8. Solve Example 6 by treating it from the aspect of energy.
Kinetic energy of 500 kg slide at 04 m/s
mv 2 _ 500 X 04 x 04 _
2 ~ 2
The slide is stopped in 1 second from a speed of 04 m/s
Average speed = 04 ~ 2 = 02 m/s
space = (velocity) (time) =02x1= 02 m
Hence 40 J of energy is dissipated over a distance of 02 m and if F is
the force acting to do it
Work = (force) (distance)
40 = F x 02 F = 40 s 02
= 200 N as before.
Example 9. A 250 kg drop stamp falls through a height of 4 m before
striking the work. If it compresses the metal and is brought to rest in a
distance of 25 mm, estimate the average force of the blow.
Work stored up in stamp when striking work
= potential energy = mgh = 250 x 981 x 4 = 98 10 J
This is dissipated over a distance of 25 mm = 0025 m
Work = (force) (distance)
9810 = F x 0025
7om 392 400N
r aQ25 j?x huui>
CIRCULAR MOTION 209
Circular motion
Many of the problems of the shop deal with rotating masses (flywheels,
pulleys, etc), so that we must adapt our knowledge of motion and force
to circular as well as linear movement.
It is quite permissible to use the equations of motion as they stand
and apply them to some point on the rotating body. In general, this will
be a point on the rim should the body in question be a flywheel or pulley.
It is better, however, when dealing with circular motion, to use
the angular notation of quantities, since if we consider any rotating body
the speed of points at different radii will vary according to their radius.
The angular speed of the body, however, is constant, irrespective of the
radius.
Fig. 152
We have seen that if we consider an angle AOB, where AB is the arc
AB
struck from the centre O, and r is the radius, then = 6 in radians.
r
If this refers to a body having circular motion about O as centre:
if arc AB = space moved (s), then  = angle (radians), or
s = Or
Again, if AB represents the velocity (v) of the body at radius r, then
 = angular velocity (&>)
or v = cor (7)
If v is in m/s, and r in m, then to will be in radian per second.
In the same way the relation between the linear acceleration a and the
angular acceleration a of a body moving in a circle is that
a
— = a or a = ar
r
210 MECHANICAL PRINCIPLES — III
When the reader has become accustomed to these simple conversions
from the linear to the angular notation, he will find the angular working
is preferable to and more rational than linear working for problems of
rotation .
The equations of motion given on pages 198 later may be used for
angular notation when it is remembered that:
s = space in radians; v = velocity in rad/s
a = acceleration in rad/s 2
Since 1 revolution = In radian we may convert rev/min to rad/s by
dividing by 60 and multiplying by 2n,
N r i
i.e. (o (rad/s) = tk^ 71  l N = rev/minJ
oU
Example 10. A pulley revolving at 220 rev/min comes to rest with uniform
retardation in 50 turns. Calculate the angular retardation and the time
taken .
We may obtain the time very simply without using radians, for if the
retardation is uniform, the velocitytime graph is a straight line and the
average velocity = —= = 1 10 rev/min
The pulley therefore comes to rest in 50 turns at a mean speed of 1 10
rev/min, i.e. in fYfc min = ffi x 60 = 27 sec.
Using the equation cv 2 = oj x + at
U = pfTMi + a. 27.
oU
220 X 2n „ _, ,. ,
a =  ~zr ^pt = °' 853 rad/s 2
Ol) X 2/
Example 11. A gas engine which normally runs at 500 rev/min takes 20 s
to accelerate its flywheels to their full speed. Calculate the angular
acceleration and the number of revolutions made by the engine. (Assume
constant acceleration.)
We may obtain the number of revolutions from the average speed as
before, since average speed = =— = 250 rev/min
250
Using s = vt we have s = ^ X 20 = 833 revolutions.
oU
ACCELERATING TORQUE 211
To find the acceleration we may use
a> 2 = a> l + at
500 on
Zrr2 71 = + a.20
ol)
500 X 2.T
a =
2 = 263 rad/s
60 x 20 6
Exercises 8c
1. Convert a speed of 175rev/min to rad/s, and llOrad/s to rev/min.
2. A pulley has an acceleration of 4 rad/s 2 . How long will it take to reach a speed of
250 rev/min?
3. A shaft retards from a speed of 500 rev/min to rest in 3 second. Calculate its retarda
tion in rad/s 2 .
4. Calculate the kinetic energy of a car of mass 1400 kg and travelling at 54km/h.
If this is brought to rest by the brakes in 30m, find the average force of resistance exerted.
5. Solve Ex. Sb, No. 7, from a consideration of energy.
6. A machine table of mass 150 kg is moving at 30m/min. Calculate its kinetic energy.
What force applied to it will bring it to rest in 600mm? (Neglect friction.)
7. Find the energy stored in a 4kg hammerhead moving at 2m/s. If this hammer is
brought to rest by compressing the metal under it through a distance of 25 mm, determine
the average force of the blow.
Accelerating torque
When a torque acts on a rotating body or on a body capable of rotation
it causes angular acceleration.
In Fig. 153 let the small mass m be rotating about the centre O at
radius r and let the force F act on it.
Then F = ma, where a = linear acceleration of m.
But torque T = Fr, and angular accel a = , i.e. a = ar.
Hence F = ma = mar and T — Fr = mar 2 .
But a rotating body is made up of many small masses m, each situated at
its own particular radius.
The sum of all these small masses we will call M, the total mass of
the body. If this mass could be concentrated at a single radius which we
will call k, we could write for the whole rotating body: T = Mk 2 a . Actually
it is possible to determine the value of a radius at which the whole of the
mass of a body may be assumed as concentrated, and this radius is called
the Radius of Gyration, being denoted by k.
212 MECHANICAL PRINCIPLES
We may thus write:
Accel torque (T) = Mk 2 a
(8)
For a flywheel having a rim heavy in proportion to the rest of the
wheel, k may be assumed as the mean radius of the rim.
For a plain disc of radius r, k = 0707 r.
We might caution the reader again regarding the units of the above
expression. If T is in newton metres (Nm), M will be in kilogrammes,
k will be in metres and a will be in rad/s 2 .
S/
I
m F
O l
s
\
/
If 60*0 25
* 40N
160N
Fig. 153
Fig. 154
Example 12. A flywheel of mass 550kg has a heavy rim, the inside and
outside radii of which are 450 mm and 550mm respectively. When this
wheel is rotating at 105 rev/min a brake is pressed against the rim with a
radial force of 160N. If the coefficient of friction between the brake and
the wheel is 025, calculate how long the wheel will take to come to rest
and how many revolutions it will make in doing so.
The tangential force at the rim of the wheel tending to stop it is
160 x fx.
= 160 x 025 = 40N
Since this acts at a radius of 550 mm = 055 m, the retarding torque will
be 40 x 055 = 22Nm.
As the rim is heavy in comparison with the rest of the wheel, we will
take the Radius of Gyration at the mean rim radius, i.e. at 500mm radius
(05 m).
ACCELERATING TORQUE 213
Using the torque equation T = Mk 2 a we have
22
22 = 550 x 05 X 05 x a from which a = 5Q Q  Q  = 016rad/s 2
We may now use the equation co 2 = w, + at to find the time to stop
the wheel since
n 105 n ^
and a = —01 6 (retardation)
Hence = ^2* 016*
i.e. In — 016?
60
t = tr n X n %* = 68 ' 8 seconds
60 x 7 x 016
Since the average speed of the wheel in coming to rest is 525 rev/ min
it will make 525 X pk~ in 688 sec, i.e . 602 re v.
6U
Example 13. A pulley is in the form of a castiron disc 500mm x 80mm
thick. It is driven from a source of power which exerts a constant torque
of ION m. How long will this pulley take to attain to a speed of 210rev/min
when started from rest? Take the density of the cast iron to be 7280 kg/m 3 .
Mass of pulley = j DH X density
= 2 x . 5 2 x . 08 x 7280 kg = 1144 kg
Taking the radius of gyration as 0707 x outside radius, we have k =
0707 X 025 = 01 72 m
The applied torque is lONm and applying the torque equation
T = Mk 2 a, 10 = 1144 X 0172 2 X a
<x = ttta nmi = 28rad/s 2
H44 x 0172 2
co = 2l0rev/min = 44rad/s
44
1 ~ S3
157 seconds
214 MECHANICAL PRINCIPLES — III
Example 14. A steel level 300mm long is of rectangular crosssection
40 mm x 20 mm, and is pivoted at its centre . One end bears on a cam which
causes it to swing through a length of 25 mm, during the time that the cam
makes \ revolution at 180rev/min. If half the swing of the lever is made
with constant acceleration, and the other half with constant retardation,
estimate the torque required to accelerate the lever. Ik for a lever pivoted
at the centre = rapprox(L = length of lever).] Take the density of steel
as 7840 kg/m 3
The problem is shown in Fig. 155.
Fig. 155
Approximate mass of lever = 03 x 004 x 002 x 7840 = 188 kg.
The lever is accelerated through an arc 25 mm long on a 150 mm
radius, i.e. through an angle of £ radian.
This takes place during £ turn of a cam revolving at 180 rev/min, i.e.
\ rev at 3 rev/s
= £ X j = 27 second
Since = <ot + \at 2 = \at 2 when co =
If) 1 v i
.. 9 = \at\ and a = =g = ^jkr = 192 rad/s 2
For the lever we have radius of gyration k = pr = — yy — =
25 mm = 0025 m, and mass m = 188 kg.
Hence applying the torque equation
T = mk 2 a = 188 x (0025) 2 X 192 = 0225Nm
This means that with the cam end of the lever being accelerated upward
a torque of the above amount would have to be applied by the cam to the
THE ENERGY OF ROTATING BODIES 215
end of the lever. With the cam end falling, this torque would have to be
applied to the lever by a spring or other means to prevent the roller from
leaving the cam.
The above consideration is, of course, relative to the lever only, and
neglects the effect of attachments to its free end.
The energy of rotating bodies
If we continue our consideration of a rotating body as having all its mass
concentrated at the radius of gyration (k), if the body is rotating at a
speed of a> radians/second the linear speed in metres/second of a point
at radius k will be
v = o) k, if k is in metres.
mv 2
Since the kinetic energy of a body = ^; for a rotating body it will be
m(a)k) 2
newton metres
Hence KE = — = — joules
Rotating flywheels are used on presses and other machines for the
purpose of storing a reserve of energy, so that when a sudden large out
put of work is required the machine will not stall, or undergo an undue
slowing up in speed.
Example 1 5 . Calculate the energy stored up in a solid disc of cast iron,
500 mm dia by 80 mm thick, when rotating at 180 rev/min. If this wheel
is on a press, what will be its speed after a hole has been blanked if the
blanking pressure is 100 kN and its duration extends over 5 mm.
From Example 13 the mass of this wheel is 1144 kg and its radius of
gyration, 01 72 m.
180
The angular speed o> = ^r x In = 6;rrad/s
oil
Putting these in the expression for Kinetic Energy
„„ mk 2 (o 2 1144 x 01 72 2 x (6tt) 2
KE =  T  = 2
Energy stored at 180 rev/min = 6015 J
216 MECHANICAL PRINCIPLES — III
For a pressure of 100 kN extending over 5 mm, the work done
= 100 000 x 0005 = 500J
This energy must be given up by the flywheel and its energy afterwards
will be 6015  500 = 1015 J
If oj x is the speed of the wheel after giving up the energy: reapplying
the expression for kinetic energy
Tm , mk*a>* 2 2 x 1015 203
1015 = —  — : u> 1 —
2 ' rnk 2 1144 X 0172 2
= 60 (jo = ^60 = 775 rad/s
co 775
rev/s = x = = —
In In
775 X 60 2325 _, , .
rev/m = = = = 74 rev/min
2.71 71
Hence speed of wheel after blanking hole = 74 rev/min
Example 16. Solve Example 12 from a consideration of energy and work
done.
At 105 rev/min the work stored up in the flywheel = Energy = — = —
550 X 05 X 05 X 11 X 11
2
= 8320 J
The tangential force at the brake block = 40 N, and to dissipate 8320 J
8320
of energy this must travel —tpt = 208 m
Revs of wheel for 208 m on its circumference
208 208 , A « , ,
602 rev, as before
Circum n x 11
Average speed = 525 rev/min
Hence time to stop = py^ X 60 min = 688 s
Example 17. A flypress has two 150mm diameter spherical castiron
balls spaced at the ends of the arm, each ball being at 500 mm radius. At
THE ENERGY OF ROTATING BODIES 217
Fig. 156
what speed must the arm rotate if the punch is to be just capable of
penetrating a thickness of 2 5 mm under a constant pressure of 20 000 N?
Take the density of cast iron to be 7280 kg/m 3 .
The reader is no doubt acquainted with the flypress, a sketch of which
is shown in Fig. 156.
mass of a ball = volume x density
= n * ai53 x 7280= 12.9kg
Energy stored in 2 balls at a speed of co rad/s.
= 2 pM ) = mk2(o2
= 129 X 05 X 05 X <o 2 = 3.225a; 2
This must be equal to the work output required at the punch, i.e.
20 000N acting for a distance of 2 5 mm = 00025 m
= 20 000 x 00025
= 50J
Hence 3225w 2 = 50,
= Vl55
50
3225
= 155,
394
394 rad/s = ^P = 0627 rev/s
218 MECHANICAL PRINCIPLES — HI
Exercises 8d
1. A CI flywheel rim is .1 m outside, 08m inside diameter and 01m wide. Taking the
radius of gyration as the mean radius, calculate the energy stored up when the speed is
315rev/min [Density of CI = 7280 kg/m 3 ].
2. If the flywheel in Ex. No. 1 is attached to a press and a punching operation causes
the speed of the wheel to drop to 2625 rev/min, calculate the energy absorbed by the
operation.
3. A CI flywheel, rim 14m outside diameter, lm inside, and 01m wide is revolving
at 105 rev/min when two brake pads are pressed against opposite sides of its rim. If the
coefficient of friction at the brakes is 04, with what force must they be pressed against the
rim to bring the wheel to rest in 30 revolutions?
4. A motor develops a constant torque of 32Nm. It drives a shafting and pulleys of
mass 400 kg, and having a radius of gyration = 200 mm. Calculate the time taken for
the motor to accelerate the shaft from rest to a speed of 3 1 5 rev/min.
5. A flywheel is in the form of a solid castiron disc 06 m diameter, 80mm thick, and is
carried on a 50mm diameter shaft. When the wheel is revolving at 315 rev/min it is dis
engaged from its drive and comes to rest in 75 seconds . If the retarding torque is due entirely
to friction at the bearings, calculate the tangential frictional force at the surface of the
50mm shaft upon which the wheel is mounted.
6. A press has a flywheel of mass 400 kg and radius of gyration = 0707 m. The press is
shearing metal bars 40mm x 20mm, for which the ultimate shear strength is 400N/mm 2 .
When a bar is being sheared the full load comes on and remains constant whilst the shear
blade moves 5 mm, after which the load drops to zero. If the above flywheel is rotating at 525
rev/min when a bar is sheared, calculate the reduction in speed caused by the work absorbed.
7. A flypress has two masses each of 5 kg attached to opposite ends of the arm at
400mm radius. When the arm is rotating at 1 rev/s a punch in theram makes contact with a
pin and drives it a distance of 5mm into a hole. Calculate the average pressure exerted
on the pin by the punch.
8. A cylindrical mixing tank mass 160kg, inside diameter lm, radius of gyration
250 mm, is rotating at 105 rev/min when 20 kg of material is tipped into it. If the material
immediately moves to the sides of the tank and arranges itself at a mean radius of 400 mm,
estimate the momentary reduction in speed caused by the extra mass.
Heat and Heat Energy
Temperature
The temperature of a substance is no indication of the amount of heat
it contains, but merely a measure of its "hotness level."
There are two scales of temperature used in this country; the Celsius
and the Kelvin. (The Celsius scale is a more modern* name for the Centi
grade scale, but the latter name is likely to persist for some time after the
adoption of the SI conventions.) The freezingboiling point limits of water
are shown in Fig. 157.
AMOUNT OF HEAT 219
373
273
100"
Boiling
Freezing
Kelvin Celsius
Fig. 157
The degrees of temperature on these systems are generally denoted °C
and K respectively. The chief calculation concerning these is the conver
sion from one to the other. It will be noted that a very simple relation
ship exists,
i.e. temperature in Kelvin = temperature in Celsius + 273. For example
20°C = 273 + 20 = 293K
and 353 K = 353  273 = 80°C
(The reader will note that the word "degree" is not used when referring
to the Kelvin scale.)
Furthermore, 1°C of temperature rise = 1 K of temperature rise. The
symbol relates to the kelvin scale of absolute temperature and goes down
to absolute zero (0) of temperature. It is employed mostly in scientific
and thermodynamic work.
Amount of heat
Heat is a form of energy, and hence the unit for a quantity of heat is the
same unit as for a quantity of energy, i.e. the joule. For large quantities
we occasionally refer to kilojoules and megajoules.The amount of energy
needed to raise unit mass of a substance a unit increase of temperature
is called the specific heat capacity of that substance. As an example,
the specific heat capacity of water is about 4200J/kg°C. Water has the
greatest specific heat capacity of any substance, so that all other sub
stances have a specific heat capacity of less than 4200 J/kg°C.
220 MECHANICAL PRINCIPLES — III
Table of Specific Heat Capacities
Specific Heat
Substance Capacity
in J/kg °C
Water 4200
Alcohol (absolute) 2930
Olive oil 1300
Petroleum 2140
Turpentine 2000
Copper 400
Cast iron 540
Aluminium 900
Lead 1300
Steel 480
Brass 395
Oak 2400
Stone generally 880
Air (at constant pressure) 1000
When a substance is heated up the heat transferred to it will be
given by
(Mass of substance)(Rise in temperature)(Specific heat capacity)
= mc (T 2  T,)
[m = mass;c = sp.ht. capacity; T x and T x = Temp limits.]
Heat calculations
Due to the fact that heat escapes so easily and so rapidly by conduction,
convection and radiation, the results of heat and temperature calculations
are not to be relied upon. It is possible, however, to obtain an approximate
idea of the state of conditions, and the work involves important principles
in which we cannot have too much practice.
Example 19. A piece of steel of mass 5kg is heated up to 700° C and
immersed in a tank 400mm square containing 250mm of water at 15°C.
The steel is left until temperature conditions become steady. Assuming
that 10% of the heat is lost, estimate the final temperature.
When the hot steel is placed in the water it will lose heat to the water,
and if no heat were lost externally the heat lost by the steel would be
HEAT CALCULATIONS 221
equal to the heat gained by the water. As it is, 90% of the heat lost by
the steel is gained by the water.
Let 7 be the final steady temperature.
Heat lost by steel = (mass)(sp. ht. cap.)(fall in temp)
= 5 X 480(700  7)
The mass of 1 cubic metre of water = 1000 kg, so the mass contained
in a tank 04 m square when the depth is 025 m
= 04 x 04 x 025 x 1000 = 40kg
Heat gained by water = 40 x 4200(7  15) (2)
Since heat gained by water = i%(heat lost by steel) we have
168000(7  15) = U5 X 480(700  7)] (3)
Multiplying out the brackets we have:
168 0007 2 520 000 = 1 512 000  2 1607
From which 170 1607 = 4 032 000
««h t 4Q32Q0Q ^ 7 or
and T= 170160 = 2H1C
Example 20. A piece of steel 1 kg in mass is taken from a furnace and
placed in 2 litres of water. Conditions are made such that the minimum
of heat escapes. If the initial temperature of the water was 20°C and the
final steady temperature of steel and water 52° C, estimate the temperature
of the steel when taken from the furnace.
If we assume no loss of heat we have
Heat gained by water = Heat lost by steel.
Let 7 = furnace temperature
Mass of 2 litres of water = 2 kg
Heat gained by water = 2 X 4200(52  20)
= 8400 x 32 = 268800 J
Heat lost by steel = (mass)(sp. ht. cap.)(7 — 52)
= 1 x 480(7  52) *= 480T  24 960
Equating: 4807 24 960 = 268 800
4807 = 293760
293760
1 480 21±±
Actually this would be a low estimate as some heat would be lost.
222 MECHANICAL PRINCIPLES — III
Exercises 8e
1. The dimensions of a workshop are 40m x 16m x 125 m and the ventilation system
is such that the air in the shop is changed twice per hour. If the temperature in the shop
is 20°C and the outside temperature 12°C, calculate the heat lost per hour by the shop
due to the two air changes. [Sp. ht. capacity of air = 1000 J/kg °C, density of air =
l3kg/m 3 .] Give the answer in megajoules.
2. A piece of steel mass 5 kg is taken from a furnace at 700° C and plunged into
8 litres (8kg) of water at 15°C. If 10% of the heat is lost, estimate the temperature of
the water when steady conditions have been reached. [Sp. ht. capacity of steel =
480J/kg °C.]
3. Water flows through a gasheated boiler at the rate of 12 litres per minute, and
its temperature is raised from 20°C to 70°C by the boiler. If the efficiency of the boiler
is 70% and the gas used yields 18MJ/m 3 when burned, calculate the gas consumption of
the boiler in cubic metres per hour.
4. A piece of steel of mass 125 kg is taken from a furnace and quickly transferred
into a tank containing 3 litres of water. The initial temperature of the water was 15°C,
and the final steady temperature was 50° C. Assume that no heat escaped during the
process, and estimate the temperature of the steel when taken from the furnace.
5. An oilcooling arrangement consists of a nest of tubes through which the oil
flows, the tubes being subjected to a circulation of cooling water on their exterior. The
oil flows through the tubes at the rate of 18 litres/min, its entering temperature is
80°C, and it is desired to cool this down to 30°C. Estimate what flow of water will
be necessary if it enters at 18°C and leaves at 28°C. [l litre oil = 09 kg, sp. ht. cap. of
oil = 2000 J/kg °C.]
6. In a certain locality the cost of gas for industrial heating was 4p per cubic metre,
and the cost of coal £1200 per tonne (= 1000kg). The heating value of the gas was
18MJ/m 3 and of the coal 25MJ/kg. Assuming an efficiency of application of 90% for the
gas and 60% for the coal, compare the relative costs of heating by gas and coal.
Heat energy
Heat is a form of energy, and when mechanical work is dissipated by fric
tion it is converted into heat. Most of our mechanical energy is derived
from heat by converting it through some form of heat engine . The relation
between heat and mechanical energy was at one time called Joule's Equi
valent, after the famous scientist Joule. The use of Joule's Equivalent is
not required when using SI units, as heat and other forms of energy use
the same unit.
Example 21. If lkg of coal when burned yields 25 MJ, estimate the mass of
coal required per hour to generate 100 kW if the overall efficiency is 10%.
We have that — — 2— = Efficiency = r^.
Input 100
HEAT ENERGY 223
Output = lOOkW, Input = lOOkW x ^ = lOOOkW
1000 kW = 1000 000 W = 1000 000 J/s = 10 6 J/s
Heat energy required per hour = 10 6 X 3600 J
Heat energy in 1 kg of coal = 25MJ = 25 X 10 6 J
Mass required = — „$ x ine = 144 kg
Example 22. A machining operation is cooled by soluble oil having a
specific heat capacity of 3000J/kg°C, flowing at the rate of 601itres/min.
If 10 kW is being absorbed at the cutting point and 80% of the heat
generated is taken away by the cooling oil, calculate its rise in tempera
ture, tl litre of coolant = 09 kg].
Assuming all the power to be dissipated as heat at the cutting point,
we have:
Heat generated per minute at the cutting point
= lOkW for one minute = 10 OOOJ/s x 60s = 600 000J
Heat taken away by coolant = & x 600 000 J = 480 000 J
Heat taken up by coolant = (mass)(sp. ht. cap.)(temp rise)
= (60 X 09) (3000) (temp rise) joules
Equating these we have:
(60 x 09 x 3000) temp rise = 480 000 J
_ . 480 000 ~ Q . or
Temp  nse = 60 x 09 x 3000 =!&£
Example 23. The surface speed of a grinding wheel is 1200m/min. The
end of a steel bar of 40mm square crosssection is pressed against the
wheel for ± min, with a force of 100N. If the coefficient of friction
between the steel and the wheel is 04, and if half of the heat generated is
absorbed by the steel, being momentarily confined to a depth of 25 mm,
estimate the temperature rise at the surface in contact with the wheel
face.
If the coefficient of friction is 04 and the radial pressure 100N, the
tangential resistance will be 100 x 04 = 40N and the work dissipated
per second = 40N x 12 ^ m = 800Nm = 800 J.
.*. Work dissipated in 15 s = 12 000 J
Heat input from work = \ x 12 000 J = 6000 J
224 MECHANICAL PRINCIPLES — III
Taking the density of steel as 7840 kg/m 3 , the layer of metal to which
this heat is assumed to be confined = volume x density.
= 004 x 004 x 00025 x 7840
= 0031 36 kg
Heat given = (mass)(sp. ht. capacity)(temp rise)
6000 = (0031 36)(480)(temp rise)
Temp rise = 480 x 6 °0°031 36 = ^C
Exercises 8f
1. A cutting operation is absorbing 2kW at the tool point. Calculate the heat being
generated in kilojoules per minute. If this operation is being cooled by oil flowing at
the rate of 10 litres per min, and if 90% of the heat is taken away by the oil, calculate its
rise in temperature, [l litre oil = 09 kg; sp. ht. cap. of oil = 2500J/kg°C.]
2. Estimate the gas consumption of an engine when it is developing 12 kW with a
thermal efficiency of 20%. Take the heat value of the gas used at 18MJ/m 3 , and express
the answer in cubic metres of gas per hour.
3. A bearing is 140mm diameter and runs at 500rev/min. It carries a load of 20 kN
and the coefficient of friction is 006. Calculate the work spent per minute in friction.
If the bearing is to be cooled by a circulation of oil [sp. ht. cap. = 1600J/kg°C], and if
the temperature rise of the oil is not to exceed 8°C, what mass of oil must flow through
the bearing per minute?
4. A flywheel of mass 250 kg and having a radius of gyration of 400 mm is rotating at
210rev/min, when it is brought to rest by a brake. Calculate the heat generated at the
brake. If the brake shoe is cast iron of mass 4kg, and it absorbs half the heat generated,
estimate its rise in temperature. [Spec. ht. cap. of CI. = 540J/kg°C.]
5. A grinding operation is absorbing 5kW at the wheel. Calculate the rate of flow
required for the cooling water if all the heat generated is taken away by the water and its
rise in temperature is not to exceed 4°C. Give the answer in litres/min.
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Appendix II
ISO Standard H Holes (Hole Basis)
Tolerance Limits for Selected Holes. (Selected from BS 4500: 1969)
(Unit = 0001 mm)
Nominal Sizes
H7
H8
H9
Hll
Over
Up to
and
includ
ing
UL
LL
UL
LL
UL
LL
UL
LL
mm
6
10
18
30
50
80
120
180
mm
10
18
30
50
80
120
180
250
15
18
21
25
30
35
40
46
22
27
33
39
46
54
63
72
36
43
52
62
74
87
100
115
90
110
130
160
190
220
250
290
UL = Upper Limit LL = Lower Limit
Appendix III
British Standard H Hole
[Limits for H6 to Hll over the Range 6mm to 180mm
Abstracted from BS 1916 (1953)1
Nominal Size
High Limit (unit + 0001mm)
Low Limit
over
to
H6
H7
H8
H9
H10
Hll
H6toHll
6
10
9
15
22
36
58
90
10
18
11
18
27
43
70
110
18
30
13
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33
52
84
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50
16
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39
62
100
160
50
80
19
30
46
74
120
190
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229
Appendix VI
The Trigonometrical Addition Formulae
It is sometimes necessary to express the trigonometrical ratios of the sum
or difference of two angles in terms of their individual ratios. The condi
tions are shown in Fig. 1 58 and the formulae are quoted without proof.
If the reader wishes to pursue the proof he can do so by consulting any
trigonometrical textbook.
(a)
(b)
Fig. 158
In Fig. 158 (a),
sin (A + B) = sin A cos B + cos A sin B
cos (A + B) = cos A cos B — sin A sin B
tan A + tanB
1  tan A tan B
tan (A + B) =
In Fig. 158(6),
sin (A — B) = sin A cos B — cos A sin B
cos (A — B) = cos A cos B + sin A sin B
, . _,. tan A — tan B
tan (A — B) =  t r— — ^
v 1 + tanA tanB
By writing B = A in (A + B) we get:
sin 2 A =2 sin A cos A
cos 2 A = cos 2 A — sin 2 A = 2 cos 2 A — 1 = 1 — 2 sin 2 A (since
tan 2 A
cos 2 A + sin 2 A = 1)
2 tanA
1  tan 2 A
230
Appendix VII
Continued Fractions
A continued fraction is a series of fractions derived from a single compli
cated fraction, each fraction in the series approaching more nearly to the
value of the fraction from which the series was derived.
If during the course of our work we required to obtain a gear ratio
131
(e.g. for cutting an obscure screw pitch) of, say, r—, we could only do
it by having gears containing 131, and 353 teeth, because neither of the
numbers has any factors. To cut such gears for the job would be out of the
question, as in the first place they would be too large to fit on to the
machine, and the cost of making them, if they were only required once,
would be prohibitive.
By applying the method of continued fractions to such a problem, it
would probably be possible to obtain a ratio in a usable form, and having
a value so close to the original, that under the circumstances it would be
acceptable.
For the purpose of explaining the method we will convert the above
ratio to a continued fraction:
Example 18. Put the numbers down as a division sum only divide the
numerator into the denominator
Now divide this into the
original divisor
Divide into the last divisor
Divide into the last divisor
(and so on every time)
131)353(2
262
91)131(1
9J_
40)91(2
80
11)40(3
33
7)11(1
7
4)7(1
4
3)4(1
3
1)3(3
3
231
232 APPENDIX VII
The quotients obtained from the continued division are
2, 1,2,3, 1, 1, 1 and 3.
These may now be written as the denominators of a continued fraction
as follows:
A
1
C
2
+
1
1
+
E
1
G
B
D
2 +
F
1
3
H
+
J
1
1
K
+
L
1
1
M
+
N
1
1
+
1
3
We now have to find what are called the "convergents" of this fraction.
In simpler terms, the convergents are approximations to the actual value
of the fraction. They are, in value, alternately too large and too small,
but each time they approach nearer to the actual value (hence the term
convergent).
1st convergent [Down to line AB] = y
2nd convergent [To line CD]
3rd convergent [To line EF]
1
2+1 3
1
1
1
l
2+1
1 + 1
2+ 1
"2 + f
2
4th convergent [to line GH]
2+1 2+1
1 + I 1 + j_
2 + J_ 2±
3
CONTINUED FRACTIONS 233
2+1 2+ I 2 + ^ ft
K
7
_ 1 10
— 27
1 +f ^
5th convergent [to line JK] = 2 + \
+ 1 2+1
1 + 1 1 + 1
2 + 1 2 + i
3 + 1
1
1 = 1 = 1 = I
2+1 2 + J_ 2 + & H
u
9
1+1 u
6th convergent [to line LM] = y + \
1 + 1
2+ 1
3 + 1
1 + 1
I
2+ 1
2 +
1
2
+
1
1
+
1
2
+ 1
3 + i
1
+ 1
2 + f
1 + 1
16
7
1
2 + 1
1
2 + 1
1
z 23
. 1 _23
• 62 ~ 62
23
1
+
7
16
23
16
7th convergent [to line NP] . (The reader should now be able to check
this for himself) = "
8th convergent [whole fraction]  m
353
The values of the convergents, their decimal equivalents and the error
(+ or ) from the true value of the original fraction, are set down in the
following table.
234 APPENDIX VII
Original Fraction = Mi = 03711
Convergent
No.
Value.
Decimal
Error
Equiv.
+ or .
05000
+ 01289
03333
00378
03750
+ 00039
03704
00007
03714
+00003
03710
00001
037113
rG.00003
03711
00000
If a graph of the error against the convergent No. is plotted it will be
somewhat like Fig. 159. It will be noticed that the error is only 00007
even at the 4th convergent, although this is a very simple fraction
<#)•
+
V.
O
u
\ 2
A^ 4 ^_
6
1
1
/ 3 Xi^ 5
Convergent number
" 7
Fig. 159
Answers to exercises
Exercises 2a
Hole Shaft Clearance
(mm) (mm) (mm)
1. 75120 74940 max 0226
75000 74894 min 0060
2. 35062 35018 max 0060
35000 35002 min 0018
(interference).
3. 20021 19980 max 0062
20000 19959 min 0020
4. 57046 57083 max 0007
57000 57053 min 0083
(interference).
5. Centre distance 12498 mm; "Not Go" diameter 12079 mm
6. 400062 mm diameter
7. Open out "Go" and "Not Go" ends by 0052 mm
8. Same fit with blocks on top limit. Additional clearance of 0017mm with pair of
blocks on bottom limits
9. Yes. Reduce "Go" diameter by 00373 mm and "Not Go" by 0036 mm
10. Max. 0271 mm: Min. 0 170mm
Exercises 2b
101 2. 104 3. 1005 4. 108
110 160 103 170
900 100 160 1600
2000 700 5000
2500
170 6. 106 7. 102 8. 105
24.00 130 150 140
50.00 1300 700 1700
10000 2500
109 10.(a) 1005 (b) 105 11. "Go" end. "Not go" end.
190 102 190 1005 105
300 190 1200 117 190
7500 600 2000 180 900
2500 600 800
2000 1000
12. 108
400
1000
235
236 ANSWERS TO EXERCISES
Exercises 2c
1. 030 mm 3. 1031 m 4. One end 4140 mm higher
5. 0° 6' too steep 6. BC = 3925 mm; CD = 29 15 mm
AD = 3590mm;>> = 1310mm;x = 2693
7. True angle = 37° 43'; error 0° 1' 8. 50009 mm; 0 108 mm
9. 7579mm. 10. 1016mm 11. 1486mm
Exercises 2d
1.35° 18' 2. 22° 20' 3. 47° 4' 4. (i) 39°46'; (ii)a = 8,6 = 16; (iii) 2501
5. 25° 50' 6. a = 4623 mm, a = 30° 18' 7. A = 152 19 mm; B = 5781 mm
8. AB = 1035mm; 29° 32'; 34° 12'
1. 40546 mm; 25°2'
Exercises 2e
2. ll67mm 3. A = 77508mm; B = 76008mm
4. (25  H) 2 = 225  ( 102 D ) 2 . 1546mm 5. d = j; 2667mm 6. 9° 12'
Exercises 2f
1. (a) 2053mm; (b) 3643mm; (c)0767in
2. Normal to axis, 589 mm diagonally across plugs, 5926 mm
3. 1009mm 4. 1294mm 5. 154mm 6. 52804mm
7. 1416 mm 8. 6405 mm
Exercises 3a
28
39
54 1
75
104
145
202
280
250
180
130 1
94
68
49
35
25
877
1260
1 1820
2630 rev/min
6
4
1 3
7
mm
Dia
4. 41 3 rev/min; 237mm, 119 mm, 60mm
mm Dia
5. 5500 rev/min
3228;
500
Exercises 3b
3551 h 3.252m/min 4. Tu " g g C g arb
rl .o .0 .
 27; 73 6. Rake 12° 10', Clearance 14° 50'
28
= —r approx
v  20
7. 261 mm; 21° 8. 15° 1
10. Rake 19° 36'; Clearance 15° 36'
9. a = 7° 18'; /5 = 22° 13'
Exercises 3c
1. 57°28' 2. 1 in 6128; 3. 29°
4. (a) 155 mm; (6) 065 mm; (c) 7° 32'
ANSWERS TO EXERCISES 237
5. 805mm; ±5' for 3°55' to 4°2' 6. 707mm; 247mm; 70° 38'
7. Angle 41° 18': Depth 695 mm Land 281 mm 8. 478 mm; l66mm
9. 2564 mm; 6122mm; 22° 03' with Horiz.
10. (a) 204mm; (b) 296 mm; (c) 11° 23'
Exercises 3d
1. *J = jjp x 6 * ; i.5009mm 2. J; Actual lead 31428 mm; Error + 00012mm
55 50 x 110 35
39
3. 4433 mm; nearest 4432 with rj ratio; 6647 mm/min
44
4. (a) 193 kW; (6)0778mm 3 /J 5.312mm; 6.10mm
7. 384/7 8. 2796 Nm (a)08786kW; (b) 2238 J/mm 3
9. 1400 rev/min; 1463Nm; 279 kW 10. 533 m; 195/?
20 40 x 25
11. Nearest Ratio = 49 = 35 ?() ; Pitch = 2041 mm
Exercises 4a
1. 669 mm; (a) 6952 mm (6) 049 mm
2. (a) w = 9435 mm; h = 61 mm; (6) w = 1000mm; A = 637 mm
3. Cutting 006 mm too shallow.
4. Cutting 017mm too deep; Tooth 012mm thin; Space 013mm wide.
5. (a) w = 1110mm, h = 598mm; (i) w = 1104mm, h = 595mm
6. 738 mm; (a) 25738mm (b) 17848 mm 7. 19346 mm
8. (a) 2363 mm; (b) 2345 mm
9. (a) 1845mm; (b) 1406 mm; (c)867mm; (rf)78mm
Exercises 4b
1. P.D. = 210mm; T.D. = 218mm; Depth = 9mm
2. w = 7808 mm; h = 40768 mm
3. T = 40T; P.D. = 200mm; T.D. = 208mm; t = 24f;
p.d. = 120mm; t.d. = 128mm; Depth = 9mm
4. 1797 mm 5. 0215°
6. (a) 22135mm; (b) 22935mm; (c) 9mm; (rf) 1910mm; (e) 63.
7. (a) 27; (b) P.D. 12471 mm, T.D. 13271 mm; (c) 678mm; (</) 42.
8. 2; P.D. = 7071 mm; T.D. = 7471 mm; lead = 222mm; 45 mm
9. Gear: 45T, 120mmP.D.; 125 mm T.D. ; Helix angle, 20° 22'; lead, 1016mm
Pinion: 30T, 80mm P.D., 85mm T.D.; Helix angle, 20° 22', lead, 677mm
Exercises 4c
1. T.D. = 4636 mm; R.D. = 3263mm; A = 9° 2'.
2. 35T; P.D. 11140mm; T.D. 1 1776 mm; Centre Dist 7570 mm; Throat rad. 1682mm;
Whole dia. 12471 mm; 2822mm [Face angle 75°.]
3. 6; A = 32° 29' 862 mm
4. Wheel: 75T; P.D.23873mm; throat dia. 24509 mm; rad. 2745 mm. Worm: 5start;
P.D.6127mm; T.D.6763mm; A = 14°34'
238 ANSWERS TO EXERCISES
5. T = 24; P. angle, 45°; Cone dist., 4242mm; Add. angle, 3°22'; Ded. angle, 4°14';
Face, 48°22'; Root, 40°46'; Whole dia., 6332 mm; Tip dist., 2823 mm.
6. Wheel: 39T; D = 195mm; Angles: P = 56° 18'; Add., 2° 27'; Ded., 3°3'; Face, 58°45';
Root, 53° 15'; Whole dia., 20055mm.
Pinion: 26f; d = 130mm; Angles: P = 33° 42'; Add., 2° 27'; Ded., 3° 3'; Face, 36° 9';
Root, 30° 39'; Whole dia. 13832mm; Cone dist., 11716mm.
7. 59 mm; 5° 58'.
8. Wheel: 24T; P.D. = 120mm; Angles: P = 73° 54'; Add.,4°34';Ded.,5°43';Wholedia,
12277 mm; Cone dist., 6245 mm.
Pinion: 18T; P.D. = 90mm; Angles: P = 46°6'; Add., 4°34'; Ded., 5°43'; Whole
dia., 9693 mm.
Exercises 5a
1. (a) 15; (6)6944mm; (c)488mm 2. 25 to 30; 3119mm.
3. 7°31'; negative. 4. 9°26'. 5. 82j° 6. 482mm.
Exercises 5b
1. 70rev/min, 189mm/min 2. 60480mm 3 , 294 kW 3. 3mm.
4. 204 min 5. Spiral mill. 6. 18 kW, 25 min, 015p.
Exercises 5c
[in the answers to Exercises 5c and 5d, whole numbers refer to complete turns of the
crank, numerators of fractions to holes, and denominators to hole circles. When the
fraction is a simple one (e.g. \) it has been left in that form. Numerators of gear ratios
are drivers, and denominators are driven gears.]
1. (a) 3i; (b) 2f; (c) Iff; (d) I A; (e) ? ; (J) f; (g) f?; (h) ®
2. (a) 3£; (b) 2tf; (c) ljf; (d) lfc (e) ft; CO ff; fe)§fc (h) j?
3. (a) If; (b) 2{f; (c) 3f; (d) 5{f; («) 7&.
4. (a) Iff; (b) 3&; (c) 3ff; (J) 8i; (<?) 15£
5. If; 3 mm
6. (a) ^ + tf ; (6) ^ + if; (c) §  A; (<*)£*
40C.c
cN  C«
8. (a) 8 holes; 20 circle; Gears, fj} X ?f; plate to turn in same direction as crank.
(b) 20 holes; 27 circle; Gears, ff x ft; plate turn same direction as crank, (c) 8 holes;
20 circle; Gears, ^; plate to turn opposite to crank, (d) 10 holes, 33 circle; Gears,
If X ft; plate to turn same direction as crank.
[Note. — Other solutions are possible to 8(a), (b), (c) and (d).]
9. 16°52' 10. 5ff;8f; lift; 14ff
Exercises 5d
1. (a) 1^; 10° 36' 26"; (b) 4f; 41° 24'; (c) 8^; 75° 45'
2. If. fl^ is nearer and could be obtained on Cincinnati.]
, , 4 ... 100 x 32 ... 100 x 24 , , 32 40 x 24
3  7 * 4  {a) :48x56 s (6) W4T ; (C) 72 ; (rf) 64lT48
ANSWERS TO EXERCISES 239
5. 40X 32 ; a = 36°48'
48 x 56
6. ft X if; 24° 1' between axes of work and cutter
7. 720mm lead; tf ratio; 19° 14' 8. Nearest = ft X Jg (based on 245mm lead)
9. 72 mm; gear ratio f; a = 46°03'
10. Gear ratio f; a = 22° 1'; lead = 25 mm
11. Lead = 60 mm; gear ratio = f ; a = 30°
UVofe. — Other solutions are possible to Nos. 9, 10 and 11.]
Exercises 5e
1. 29°58' 2. 67°23' 3. 49°6' 4. (a) 41°6'; (b) 97°2'
5. 554 mm; 2458 mm; 12°32' 6. (a) 50°46'; (6) 104°28'
Exercises 6a
[in the solutions to Ex. 6a the angle given is that made with the vertical or horizontal
drawn to the commencing end of the vector.]
2. (a) 721 upwards; 34° to R of vert
(b) 1085 downwards to L; 41° below horiz
(c) 1094 downwards; 4° to R of vert
(d) 31 R to L; 44° below horiz
(e) 34 downwards; 13° to L of vert
if) 12 L to R; 5° below horiz
3. (a) 72 upwards; 34° to L of vert
(b) 1085 upwards; 41° above horiz
(c) 22 upwards; 21° to R of vert
(d) 134 R to L; 32° above horiz
4. (<?) 145 upwards; 50° to R of vert
(/) 93 downwards; 12° to L of vert
5. ac — 66; cb = 55; abc = 50°.
6. 5 units L to R; 37° above horiz
7. ac = 21; cZ> = 76 vert downwards.
8. ab = 87 L to R; 30° above horiz; be = 5 downwards; 30° to R of vert
Exercises 6b
1. R to L (outwards); 11° 18' to lathe centre line. 2. Horiz 383N; vert 321 N
3. 212N; 53N/mm 2 4. 0085m/s; 45° to horiz. 5. 2663N
6. Suspension chain 196N; pulling chain 100N 7. 233N; 552N
Exercises 6c
1. 1 530N in a plane at 34° to vertical; line of action inclined at 71° to 500N force.
2. 11mm 3. 309N; 904N 4. 216N.; 34° to horiz.
5. 130N; 200N 6. 520N; 26° to horiz. 7. 1635N horiz. L. to R.
8. 2860N 9. 274mm rad; 100° from 50 kg mass 10. 5550N
11. 765cm 3 ; 113° from A
240 ANSWERS TO EXERCISES
Exercises 6d
1. (a) 536 m/s,; (b) 464 m/s 2. 084 m/s
4. l3m/s, 0775mm 3 5. 017m/s; 23° to OA
7. 0088 m/s 8. 856N
10. (a) f; (6) 0244m/s. 11. 000052m/s
3. (a) 114; (b) 168 rpm.
6. 08 m/s
9. 00051 m/s; 7840 N
Exercises 6e
1. 170N 2. 8N 3. Headstock 319N; tailstock 531N
4. Front bearing 1620N upwards; rear bearing 417N downwards.
5. Right 357N; left, 340N 6. 1600N, 600N
7. 177N 8. 80N
1.
4.
256N; 254N at 6° 57'
270N
to horiz.
Exercises 7a
2. 750N
5. 01184kW
3.
6.
12 300N
0792 kW
1.
5.
9.
364; 939N 2.
685%; 18 060N 6.
2340J, 3857N 10.
l97kW
121N
1200N
Exercise 7b
3. 519N
7. 377; 53N
11. lONm
Exercises 7c
4. 118N
8. 351%
1. d = 447mm; L = 894mm
2. RH load = 1080N; B. press = 027 N/mm 2
LH load = 1170N; B. press = 029 N/mm 2
3. n = 00158, 522 J 4. 862 mm
6. 50 300N 7. 7950N 632N
5. fi = 0032, 114kW
8. 509N/mm 2 ; 0204mm
9. 231 N/mm 2
1. 10rev/min 2 ; 75 revs
4. 144 m/s 2
7. lm;5s
10. Stress = 200N/mm 2 ; tension = 40 000N
Exercises 8a
2. 128 s, 125 m/s 3. 0333 m/s 2 ; 12 s
5. 04m/s 2 ; 06m/s 6. 180rev/min 2 ; 1 min
8. 05m/s; 15s
Exercises 8b
1. ION 2. lm/s 3. 355N
5. 173N 6. l28m/s 2 ; 192N; 096m/s; 01849kW
7. 108m/s; 5890m/s 2 ; 589kN 8. 320mm
4. 72N
1. 1831; 1050.
4. 157 500 J; 5250N
1. 22 700 J
5. 131N
Exercises 8c
2. 655 s
6. 1875J; 3125N
Exercises 8d
2. 6940 J 3. 101 N
6. 165rev/min 7. 6320N
3. 149rad/s 2
7. 8 J; 320 J
4. 165 s
8. 136rev/min
ANSWERS TO EXERCISES 241
1. 1664 MJ
4. 785°C
1. 120kJ; 48°C
4. 9680J; 224°C
Exercises 8e
2.
59°C
3.
12m 3
5.
3861itres/min
Exercises 8f
6.
Cost gas 309
Cost coal — 1
2.
12m 3 /h
3.
264 000 J; 206kg
Index
Acceleration, 198
due to gravity, 200
Angles, cutting tool, 63
(solid), evaluation of, 142
Angular indexing, 130
measurement with sine bar, 22
surfaces, location of points on, 32
velocity and acceleration, 209
Arithmetical progression, 57
Backlash, 95
Balancing of masses in one plane, 159
Bar (pressure), 5
Bearing friction, 192
pressure, 191
Bevel gearing, 106
Bores, measuring large bores, 28
BSI Limit System, 14
Cam milling, 138
Celsius temperature, 218
Centigrade temperature, 218
Change wheels (odd threads), 77
Circular motion, 209
Coefficient of friction, 177
Continued fraction, 231
Conventions SI, 4
Cosecant, 92
Cotangent, 92
Cutters (milling), clearance, 117
fluting angle, 115
grinding, 117
number of teeth, 113
speeds and feeds, 121
tooth rake, 114
Cutting power, 79
speed range, 58
tool angles, 63
forces acting on, 157
life, 61
Dividing head, 123
Drilling power, 8 1
Efficiency of a machine, 181
Energy, 206
heat, 218, 222
of rotating bodies, 2 1 5
Equations, of motion, 198
Equilibrium, conditions for, 155, 171
Faceplate, balancing of work on, 159
Feeds for milling, 121
Fluting angle for milling cutter teeth,
115
Force, 203
of hammer blow, 205
Forces, acting on a cutting tool, 1 57
in a mechanism, 167
vectorial representation, 1 5 1
Form tools, 68
Fraction, continued, 231
Friction, 176
at a bearing 192
at nut face, 186
coefficient, 177
in screw thread, 182
of sliding keys, 188
when clamping, 178
Gauge, point for bores, 28
slip, 17
Gauging large radii, 25
Gear tooth form, 85
rack, 86
vernier, 87
Gearing, backlash, 95
base pitch, 92
bevel, 106
helical, 96
plug method of checking, 90
242
INDEX 243
stub, 94
worm, 102
Geometric speed range, 58
Graph of motion, 10, 199
Hammer blow, force of, 205
Heat, 218
of work, 222
specific capacity, 219
Helical gearing, 96
Helix, lead and angle, 98
Inclined plane, 182
Indexing, angular, 130
compound, 126
differential, 128
simple, 125
Involute, 84
ISO Limit system, 14
ISO Metric thread, 48
Johannson gauges, 17
Kelvin, 219
Kilogramme, 6
Level (spirit), 19
Limit, 13
systems, 14
Lines on angular surfaces, true length, 32
Mass, 6, 203
Measurement of gears, 8790
by slip gauges, 17
of large bores, 28
of large radii, 25
of tapers, 23, 38,
Mechanical advantage, 181
Metre, 2
Milling cam, 138
cutters, 113,
feeds, 121
fluting angle, 115
indexing, 125 et seq
power, 220
spiral, 134
Modulus of Elasticity, 195
Moment of a force, 170
Momentum, 203
Motion in a circle, 209
equations of, 198
Newton (force), 3
Pitch (base) of gears, 92
Power for cutting, 79, 122
Progression, arithmetic and geometric,
57
Rack (gear), 86
Radian, 209
Rake on milling cutter teeth, 1 14
Screw cutting, calculations for odd
threads, 77
thread as inclined plane, 182
threads, measurement by wires, 47
Secant, 92
Sine bar, 22
SI units, 3
Slip gauges, 17
Specific heat capacity, 219
Speed range geometric, 58
Spiral milling, 134
Spirit level, 19
Strain, 193
Stress, 193
Stub teeth, 94
Taper, measurement with balls and
rollers, 38
sine bar, 23
turning, 66
Temperature, 218
Tolerance, 13
Tool angles, 63
life calculations, 61
Tools, form, 68
Torque to cause acceleration, 211
Trigonometrical addition formulae, 230
Turning moment, 211
Units SI metric, 3
Vector diagrams of velocity, 162
244 INDEX
Vectors, 148
addition and subtraction, 148
applications, 151
Velocity ratio of a machine, 181
vectorial representation, 162
Vernier (gear tooth) calculations, 87
Whitworth thread, 48
Wire measurement of screw threads, 47
Work, 206
Worm gearing, 102
Young's Modulus, 195
'This is a new edition with a new format, and completely revised to present its
material in SI units; the introductory chapter leads straight into the "new"
units and includes some worked exercises with them so the reader is left in no
doubt as to what is to follow.
'The book divides neatly into two parts, the first dealing with machine tool
calculations, liberally spaced in the text and very clearly illustrated. Each
section is concluded by a number of problems to be solved by the student;
answers are supplied at the end of the text.
"The second part of the book deals with mechanic's principles, including
statics, dynamics and a few pages on heat, invariably a product of a
machine operation. The text concentrates its attention on the practical
aspects of the principles involved and rightly leaves the more academic con
siderations to other authors and other places.
'The author and his books on workshop technology have enjoyed great
popularity for 30 years, the books being standard works for many technical
colleges: it is easy to see the reason for the appeal, for this edition is neat
and compact yet the material is not overcompressed. We can expect to see it
in use for another 30 years.'
Marine Engineers Review
o
Other books in SI units by Dr Chapman:
Elementary Workshop Calculations
Workshop Technology Part 1, fifth edition
Workshop Technology Part 2, fourth edition
Workshop Technology Part 3, third edition
EDWARD ARNOLD
£2.50 net
ISBN: 7131 3260 4